DIFFERENTIAL GEOMETRY: AN OVERVIEW
I am coming more and more to the conviction that the necessity of our geometry cannot be demonstrated, at least neither by, nor for, the human intellect. . . . geometry should be ranked, not with arithmetic, which is purely aprioristic, but with mechanics.
(1817)
We must confess in all humility that, while number is a product of our mind alone, space has a reality beyond the mind whose rules we cannot completely prescribe. (1830)

§8.1. AN OVERVIEW OF PART III

Gravitation is a manifestation of spacetime curvature, and that curvature shows up in the deviation of one geodesic from a nearby geodesic ("relative acceleration of test particles"). The central issue of this part of the book is clear: How can one quantify the "separation," and the "rate of change" of "separation," of two "geodesics" in "curved" spacetime? A clear, precise answer requires new concepts.
"Separation" between geodesics will mean "vector." But the concept of vector as employed in flat Lorentz spacetime (a bilocal object: point for head and point for tail) must be sharpened up into the local concept of tangent vector, when one passes to curved spacetime. Chapter 9 does the sharpening. It also reveals how the passage to curved spacetime affects 1 -forms and tensors.
It takes one tool (vectors in curved geometry, Chapter 9) to define "separation" clearly as a vector; it takes another tool (parallel transport in curved spacetime, Chapter 10) to compare separation vectors at neighboring points and to define the "rate of change of separation." No transport, no comparison; no comparison, no meaning to the term "rate of change"! The notion of parallel transport founds itself

Concepts to be developed in Part III:

Tangent vector
Geodesic
Covariant derivative
Geodesic deviation
Spacetime curvature
This chapter: a Track-1 overview of differential geometry
on the idea of "geodesic," the world line of a freely falling particle. The special mathematical properties of a geodesic are explored in Chapter 10. That chapter uses geodesics to define parallel transport, uses parallel transport to define covariant derivative, and-completing the circle-uses covariant derivative to describe geodesics.
Chapter 11 faces up to the central issue: geodesic deviation ("rate of change of separation vector between two geodesics"), and its use in defining the curvature of spacetime.
But to define curvature is not enough. The man who would understand gravity deeply must also see curvature at work, producing relative accelerations of particles in Newtonian spacetime (Chapter 12); he must learn how, in Einstein spacetime, distances (metric) determine completely the curvature and the law of parallel transport (Chapter 13); he must be the master of powerful tools for computing curvature (Chapter 14); and he must grasp the geometric significance of the algebraic and differential symmetries of curvature (Chapter 15).
Unfortunately, such deep understanding requires time-far more time than one can afford in a ten-week or fifteen-week course, far more than a lone reader may wish to spend on first passage through the book. For the man who must rush on rapidly, this chapter contains a "Track-1" overview of the essential mathematical tools ( § 88.4 8.7 § 88.4 8.7 §88.4-8.7\S 88.4-8.7§88.48.7 ). From it one can gain an adequate, but not deep, understanding of spacetime curvature, of tidal gravitational forces, and of the mathematics of curved spacetime. This overview is also intended for the Track-2 reader; it will give him a taste of what is to come. The ambitious reader may also wish to consult other introductions to differential geometry (see Box 8.1).

Box 8.1 BOOKS ON DIFFERENTIAL GEOMETRY

There are several mathematics texts that may be consulted for a more detailed and extensive discussion of modern differential geometry along the line taken here. Bishop and Goldberg (1968) is the no. 1 reference. Hicks (1965) could be chosen as a current standard graduate-level text, with O'Neill (1966) at the undergraduate level introducing many of the same topics without presuming that the reader finds easy and obvious the current style in which pure mathematicians think and write. Auslander and MacKenzie (1963) at a somewhat more advanced level also allow for the reader to whom differential equations are more
familiar than homomorphisms. Willmore (1959) is easy to read but presents no challenge, and leads to little progress in adapting to the style of current mathematics. Trautman (1965) and Misner (1964a, 1969a) are introductions somewhat similar to ours, except for deemphasis of pictures; like ours, they are aimed at the student of relativity. Flanders (1963) is easy and useful as an introduction to exterior differential forms; it also gives examples of their application to a wide variety of topics in physics and engineering.

§8.2. TRACK 1 VERSUS TRACK 2: DIFFERENCE IN OUTLOOK AND POWER

Nothing is more wonderful about the relation between Einstein's theory of gravity and Newton's theory than this, as discovered by Élie Cartan ( 1923 , 1924 ) ( 1923 , 1924 ) (1923,1924)(1923,1924)(1923,1924) : that both theories lend themselves to description in terms of curvature; that in both this curvature is governed by the density of mass-energy; and that this curvature allows itself to be defined and measured without any use of or reference to any concept of metric. The difference between the two theories shows itself up in this: Einstein's theory in the end (or in the beginning, depending upon how one presents it!) does define an interval between every event and every nearby event; Newton's theory not only does not, but even says that any attempt to talk of spacetime intervals violates Newton's laws. This being the case, Track 2 will forego for a time (Chapters 9-12) any use of a spacetime metric ("Einstein interval"). It will extract everything possible from a metric-free description of spacetime curvature (all of Newton's theory; important parts of Einstein's theory).
Geodesic deviation is a measurer and definer of curvature, but the onlooker is forbidden to reduce a vector description of separation to a numerical measure of distance (no metric at this stage of the analysis): what an impossible situation! Nevertheless, that is exactly the situation with which Chapters 9 12 9 12 9-129-12912 will concern themselves: how to do geometry without a metric. Speaking physically, one will overlook at this stage the fact that the geometry of the physical world is always and everywhere locally Lorentz, and endowed with a light cone, but one will exploit to the fullest the Galileo-Einstein principle of equivalence: in any given locality one can find a frame of reference in which every neutral test particle, whatever its velocity, is free of acceleration. The tracks of these neutral test particles define the geodesics of the geometry. These geodesics provide tools with which one can do much: define parallel transport (Chapter 10), define covariant derivative (Chapter 10), quantify geodesic deviation (Chapter 11), define spacetime curvature (Chapter 11), and explore Newtonian gravity (Chapter 12). Only after this full exploitation of metric-free geodesics will Track 2 admit the Einstein metric back into the scene (Chapters 13-15).
But to forego use of the metric is a luxury which Track 1 can ill afford; too little time would be left for relativistic stars, cosmology, black holes, gravitational waves, experimental tests, and the dynamics of geometry. Therefore, the Track-1 overview in this chapter keeps the Einstein metric throughout. But in doing so, it pays a heavy price: (1) no possibility of seeing curvature at work in Newtonian spacetime (Chapter 12); (2) no possibility of comparing and contrasting the geometric structures of Newtonian spacetime (Chapter 12) and Einstein spacetime (Chapter 13), and hence no possibility of grasping fully the Newtonian-based motivation for the Einstein field equations (Chapter 17); (3) no possibility of understanding fully the mathematical interrelationships of "geodesic," "parallel transport," "covariant derivative," "curvature," and "metric" (Chapters 9, 10, 11, 13); (4) no possibility of introducing the mathematical subjects "differential topology" (geometry without metric or covariant
Preview of Track-2 differential geometry
What the Track-1 reader will miss
derivative, Chapter 9) and "affine geometry" (geometry with covariant derivative but no metric, Chapters 10 and 11), subjects which find major application in modern analytical mechanics [see, e.g., Arnold and Avez (1968); also exercise 4.11 of this book], in Lie group theory with its deep implications for elementary particle physics [see, e.g., Hermann (1966); also exercises 9.12, 9.13, 10.16, and 11.12 of this book], in the theory and solution of partial differential equations [see, e.g., Sternberg (1969)], and, of course, in gravitation theory.

§8.3. THREE ASPECTS OF GEOMETRY: PICTORIAL, ABSTRACT, COMPONENT

Gain the power in § 8.4 § 8.4 §8.4\S 8.4§8.4 and Chapter 9 to discuss tangent vectors, 1 -forms, tensors in curved spacetime; gain the power in § 8.5 § 8.5 §8.5\S 8.5§8.5 and Chapter 10 to parallel-transport vectors, to differentiate them, to discuss geodesics; use this power in $ 8.7 $ 8.7 $8.7\$ 8.7$8.7 and Chapter 11 to discuss geodesic deviation, to define curvature; .... But full power this will be only if it can be exercised in three ways: in pictures, in abstract notation, and in component notation (Box 8.3). Élie Cartan (Box 8.2) gave new insight into both
Geometry from three viewpoints: pictorial, abstract, component
Box 8.2 ÉLIE CARTAN, 1869-1951
Élie Cartan is a most remarkable figure in recent mathematical history. One learns from his obituary [Chern and Chevalley (1952)] that he was born a blacksmith's son in southern France, and proved the value of government scholarship aid by rising through the system to a professorship at the Sorbonne in 1912 when he was 43 . At the age of 32
he invented the exterior derivative [Cartan (1901)], which he used then mostly in differential equations and the theory of Lie groups, where he had already made significant contributions. He was about fifty when he began applying it to geometry, and sixty before Riemannian geometry specifically was the object of this research, including his text [Cartan (1928)], which is still reprinted and worth studying. Although universally recognized, his work did not find responsive readers until he neared retirement around 1940, when the "Bourbaki" generation of French mathematicians began to provide a conceptual framework for (among other things) Cartan's insights and methods. This made Cartan communicable and teachable as his own writings never were, so that by the time of his death at 82 in 1951 his influence was obviously dominating the revolutions then in full swing in all the fields (Lie groups, differential equations, and differential geometry) in which he had primarily worked.
The modern, abstract, coordinate-free approach to geometry, which is used extensively in this book, is due largely to Élie Cartan. He also discovered the geometric approach to Newtonian gravity that is developed and exploited in Chapter 12.

Box 8.3 THREE LEVELS OF DIFFERENTIAL GEOMETRY

(1) Purely pictorial treatment of geometry:
tangent vector is conceived in terms of the separation of two points in the limit in which the points are indefinitely close;
vectors are added and subtracted locally as in flat space;
vectors at distinct points are compared by parallel transport from one point to another;
this parallel transport is accomplished by a "Schild's ladder construction" of geodesics (Box 10.2);
diagrams, yes; algebra, no;
it is tied conceptually as closely as possible to the world of test particles and measurements.
(2) Abstract differential geometry:
treats a tangent vector as existing in its own right, without necessity to give its breakdown into components,
A = A 0 e 0 + A 1 e 1 + A 2 e 2 + A 3 e 3 A = A 0 e 0 + A 1 e 1 + A 2 e 2 + A 3 e 3 A=A^(0)e_(0)+A^(1)e_(1)+A^(2)e_(2)+A^(3)e_(3)\boldsymbol{A}=A^{0} \boldsymbol{e}_{0}+A^{1} \boldsymbol{e}_{1}+A^{2} \boldsymbol{e}_{2}+A^{3} \boldsymbol{e}_{3}A=A0e0+A1e1+A2e2+A3e3
just as one is accustomed nowadays in electromagnetism to treat the electric vector E E E\boldsymbol{E}E, without having to write out its components; uses a similar approach to differentiation (compare gradient operator grad\boldsymbol{\nabla} of elementary vector analysis, as distinguished from coor-dinate-dependent pieces of such an operator, such as / x , / y / x , / y del//del x,del//del y\partial / \partial x, \partial / \partial y/x,/y, etc.);
is the quickest, simplest mathematical scheme one knows to derive general results in differential geometry.
(3) Differential geometry as expressed in the language of components:
is indispensible in programming large parts of general relativity for a computer;
is convenient or necessary or both when one is dealing even at the level of elementary algebra with the most simple applications of relativity, from the expansion of the Friedmann universe to the curvature around a static center of attraction.
Newtonian gravity (Chapter 12) and the central geometric simplicity of Einstein's field equations (Chapter 15), because he had full command of all three methods of doing differential geometry. Today, no one has full power to communicate with others about the subject who cannot express himself in all three languages. Hence the interplay between the three forms of expression in what follows.
It is not new to go back and forth between the three languages, as witnesses the textbook treatment of the velocity and acceleration of a planet in Kepler motion around the sun. The velocity is written
(8.1) v = v r ^ e r ^ + v ϕ ^ e ϕ ^ (8.1) v = v r ^ e r ^ + v ϕ ^ e ϕ ^ {:(8.1)v=v^( hat(r))e_( hat(r))+v^( hat(phi))e_( hat(phi)):}\begin{equation*} \boldsymbol{v}=v^{\hat{r}} \boldsymbol{e}_{\hat{r}}+v^{\hat{\phi}} \boldsymbol{e}_{\hat{\phi}} \tag{8.1} \end{equation*}(8.1)v=vr^er^+vϕ^eϕ^
(The hats " " on e r ^ e r ^ e_( hat(r))\boldsymbol{e}_{\hat{r}}er^ and e ϕ ^ e ϕ ^ e_( hat(phi))\boldsymbol{e}_{\hat{\phi}}eϕ^ signify that these are unit vectors.) The acceleration is
(8.2) a = d v d t = d v r ^ d t e r ^ + d v ϕ ^ d t e ϕ ^ + v r ^ d e r ^ d t + v ϕ ^ d e ϕ ^ d t (8.2) a = d v d t = d v r ^ d t e r ^ + d v ϕ ^ d t e ϕ ^ + v r ^ d e r ^ d t + v ϕ ^ d e ϕ ^ d t {:(8.2)a=(dv)/(dt)=(dv^( hat(r)))/(dt)e_( hat(r))+(dv^( hat(phi)))/(dt)e_( hat(phi))+v^( hat(r))(de_( hat(r)))/(dt)+v^( hat(phi))(de_( hat(phi)))/(dt):}\begin{equation*} \boldsymbol{a}=\frac{d \boldsymbol{v}}{d t}=\frac{d v^{\hat{r}}}{d t} \boldsymbol{e}_{\hat{r}}+\frac{d v^{\hat{\phi}}}{d t} \boldsymbol{e}_{\hat{\phi}}+v^{\hat{r}} \frac{d \boldsymbol{e}_{\hat{r}}}{d t}+v^{\hat{\phi}} \frac{d \boldsymbol{e}_{\hat{\phi}}}{d t} \tag{8.2} \end{equation*}(8.2)a=dvdt=dvr^dter^+dvϕ^dteϕ^+vr^der^dt+vϕ^deϕ^dt
Planetary orbit as example of three viewpoints
Figure 8.1.
A Keplerian orbit in the sun's gravitational field, as treated using the standard Euclidean-space version of Newtonian gravity. The basis vectors themselves change from point to point along the orbit [equations (8.3)]. This figure illustrates the pictorial aspect of differential geometry. Later (exercise 8.5) it will illustrate the concepts of "covariant derivative" and "connection coefficients."
The unit vectors are turning (Figure 8.1) with the angular velocity ω = d ϕ / d t ω = d ϕ / d t omega=d phi//dt\omega=d \phi / d tω=dϕ/dt; so
d e r ^ d t = ω e ϕ ^ = d ϕ ^ d t e ϕ ^ (8.3) d e ϕ ^ d t = ω e r ^ = d ϕ d t e r ^ . d e r ^ d t = ω e ϕ ^ = d ϕ ^ d t e ϕ ^ (8.3) d e ϕ ^ d t = ω e r ^ = d ϕ d t e r ^ . {:[(de_( hat(r)))/(dt)=omegae_( hat(phi))=(d( hat(phi)))/(dt)e_( hat(phi))],[(8.3)(de_( hat(phi)))/(dt)=-omegae_( hat(r))=-(d phi)/(dt)e_( hat(r)).]:}\begin{align*} & \frac{d e_{\hat{r}}}{d t}=\omega e_{\hat{\phi}}=\frac{d \hat{\phi}}{d t} e_{\hat{\phi}} \\ & \frac{d e_{\hat{\phi}}}{d t}=-\omega e_{\hat{r}}=-\frac{d \phi}{d t} e_{\hat{r}} . \tag{8.3} \end{align*}der^dt=ωeϕ^=dϕ^dteϕ^(8.3)deϕ^dt=ωer^=dϕdter^.
Thus the components of the acceleration have the values
(8.4a) a r ^ = d v r ^ d t v ϕ ^ d ϕ d t = d 2 r d t 2 r ( d ϕ d t ) 2 (8.4a) a r ^ = d v r ^ d t v ϕ ^ d ϕ d t = d 2 r d t 2 r d ϕ d t 2 {:(8.4a)a^( hat(r))=(dv^( hat(r)))/(dt)-v^( hat(phi))(d phi)/(dt)=(d^(2)r)/(dt^(2))-r((d phi)/(dt))^(2):}\begin{equation*} a^{\hat{r}}=\frac{d v^{\hat{r}}}{d t}-v^{\hat{\phi}} \frac{d \phi}{d t}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \phi}{d t}\right)^{2} \tag{8.4a} \end{equation*}(8.4a)ar^=dvr^dtvϕ^dϕdt=d2rdt2r(dϕdt)2
and
(8.4b) a ϕ ^ = d v ϕ ^ d t + v r ^ d ϕ d t = d d t ( r d ϕ d t ) + d r d t d ϕ d t . (8.4b) a ϕ ^ = d v ϕ ^ d t + v r ^ d ϕ d t = d d t r d ϕ d t + d r d t d ϕ d t . {:(8.4b)a^( hat(phi))=(dv^( hat(phi)))/(dt)+v^( hat(r))(d phi)/(dt)=(d)/(dt)(r(d phi)/(dt))+(dr)/(dt)(d phi)/(dt).:}\begin{equation*} a^{\hat{\phi}}=\frac{d v^{\hat{\phi}}}{d t}+v^{\hat{r}} \frac{d \phi}{d t}=\frac{d}{d t}\left(r \frac{d \phi}{d t}\right)+\frac{d r}{d t} \frac{d \phi}{d t} . \tag{8.4b} \end{equation*}(8.4b)aϕ^=dvϕ^dt+vr^dϕdt=ddt(rdϕdt)+drdtdϕdt.
Here is the acceleration in the language of components; a a a\boldsymbol{a}a was the acceleration in abstract language; and Figure 8.1 shows the acceleration as an arrow. Each of these three languages will receive its natural generalization in the coming sections and chapters from two-dimensional flat space (with curvilinear coordinates) to four-dimensional curved spacetime, and from spacetime to more general manifolds (see § 9.7 § 9.7 §9.7\S 9.7§9.7 on manifolds).
Turn now to the Track-1 overview of differential geometry.

§§8.4. TENSOR ALGEBRA IN CURVED SPACETIME

To see spacetime curvature at work, examine tidal gravitational forces (geodesic deviation); and to probe these forces, make measurements in a finite-sized laboratory. Squeeze the laboratory to infinitesimal size; all effects of spacetime curvature become infinitesimal; the physicist cannot tell whether he is in flat spacetime or curved spacetime. Neither can the mathematician, in the limit as his domain of attention squeezes down to a single event, P 0 P 0 P_(0)\mathscr{P}_{0}P0.
At the event P o P o P_(o)\mathscr{P}_{o}Po (in the infinitesimal laboratory) both physicist and mathematician can talk of vectors, of 1 -forms, of tensors; and no amount of spacetime curvature can force the discussion to change from its flat-space form. A particle at P o P o P_(o)\mathscr{P}_{o}Po has a 4 -momentum p p p\boldsymbol{p}p, with squared length
p 2 = p p = g ( p , p ) = m 2 p 2 = p p = g ( p , p ) = m 2 p^(2)=p*p=g(p,p)=-m^(2)\boldsymbol{p}^{2}=\boldsymbol{p} \cdot \boldsymbol{p}=\boldsymbol{g}(\boldsymbol{p}, \boldsymbol{p})=-m^{2}p2=pp=g(p,p)=m2
The squared length, as always, is calculated by inserting p p p\boldsymbol{p}p into both slots of a linear machine, the metric g g g\boldsymbol{g}g at P 0 P 0 P_(0)\mathscr{P}_{0}P0. The particle also has a 4-acceleration at P o P o P_(o)\mathscr{P}_{o}Po; and, if the particle is charged and freely moving, then a a a\boldsymbol{a}a is produced by the electromagnetic field tensor F F F\boldsymbol{F}F :
m a = e F ( , u ) m a = e F ( , u ) ma=eF(dots,u)m \boldsymbol{a}=e \boldsymbol{F}(\ldots, \boldsymbol{u})ma=eF(,u)
In no way can curvature affect such local, coordinate-free, geometric relations. And in no way can it prevent one from introducing a local Lorentz frame at P o P o P_(o)\mathscr{P}_{o}Po, and from performing standard, flat-space index manipulations in it:
p = p α e α , p 2 = p α p β η α β = p α p α , m a α = e F α β u β p = p α e α , p 2 = p α p β η α β = p α p α , m a α = e F α β u β p=p^(alpha)e_(alpha),quadp^(2)=p^(alpha)p^(beta)eta_(alpha beta)=p^(alpha)p_(alpha),quad ma^(alpha)=eF^(alpha beta)u_(beta)\boldsymbol{p}=p^{\alpha} \boldsymbol{e}_{\alpha}, \quad \boldsymbol{p}^{2}=p^{\alpha} p^{\beta} \eta_{\alpha \beta}=p^{\alpha} p_{\alpha}, \quad m a^{\alpha}=e F^{\alpha \beta} u_{\beta}p=pαeα,p2=pαpβηαβ=pαpα,maα=eFαβuβ
But local Lorentz frames are not enough for the man who would calculate in curved spacetime. Non-Lorentz frames (nonorthonormal basis vectors { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} ) often simplify calculations. Fortunately, no effort at all is required to master the rules of "index mechanics" in an arbitrary basis at a fixed event P 0 P 0 P_(0)\mathscr{P}_{0}P0. The rules are identical to those in flat spacetime, except that (1) the covariant Lorentz components η α β η α β eta_(alpha beta)\eta_{\alpha \beta}ηαβ of the metric are replaced by
(8.5) g α β e α e β g ( e α , e β ) (8.5) g α β e α e β g e α , e β {:(8.5)g_(alpha beta)-=e_(alpha)*e_(beta)-=g(e_(alpha),e_(beta)):}\begin{equation*} g_{\alpha \beta} \equiv \boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta} \equiv \boldsymbol{g}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) \tag{8.5} \end{equation*}(8.5)gαβeαeβg(eα,eβ)
(2) the contravariant components η α β η α β eta^(alpha beta)\eta^{\alpha \beta}ηαβ are replaced by g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ, where
(8.6) g α β g α β 1 (matrix inverse) (8.6) g α β g α β 1  (matrix inverse)  {:(8.6)||g^(alpha beta)||-=||g_(alpha beta)||^(-1)" (matrix inverse) ":}\begin{equation*} \left\|g^{\alpha \beta}\right\| \equiv\left\|g_{\alpha \beta}\right\|^{-1} \text { (matrix inverse) } \tag{8.6} \end{equation*}(8.6)gαβgαβ1 (matrix inverse) 
i.e.,
(8.6') g α β g β γ = δ α γ ; (8.6') g α β g β γ = δ α γ ; {:(8.6')g_(alpha beta)g^(beta gamma)=delta_(alpha)^(gamma);:}\begin{equation*} g_{\alpha \beta} g^{\beta \gamma}=\delta_{\alpha}^{\gamma} ; \tag{8.6'} \end{equation*}(8.6')gαβgβγ=δαγ;
(3) the Lorentz transformation matrix Λ α β Λ α β ||Lambda^(alpha)_(beta)||\left\|\Lambda^{\alpha}{ }_{\beta}\right\|Λαβ and its inverse Λ β α Λ β α ||Lambda^(beta)_(alpha)||\left\|\Lambda^{\beta}{ }_{\alpha}\right\|Λβα are replaced by an arbitrary but nonsingular transformation matrix L α β L α β ||L^(alpha^('))_(beta)||\left\|L^{\alpha^{\prime}}{ }_{\beta}\right\|Lαβ and its inverse L β α L β α ||L^(beta)_(alpha)||\left\|L^{\beta}{ }_{\alpha}\right\|Lβα :
(8.7) e β = e α L α β , p β = L β α p α , (8.8) L β α = L α β 1 ; (8.7) e β = e α L α β , p β = L β α p α , (8.8) L β α = L α β 1 ; {:[(8.7)e_(beta)=e_(alpha^('))L^(alpha^('))_(beta)","quadp^(beta)=L^(beta)_(alpha)^(')p^(alpha^('))","],[(8.8)||L^(beta)_(alpha^('))||=||L^(alpha)_(beta)||^(-1);]:}\begin{gather*} \boldsymbol{e}_{\beta}=\boldsymbol{e}_{\alpha^{\prime}} L^{\alpha^{\prime}}{ }_{\beta}, \quad p^{\beta}=L^{\beta}{ }_{\alpha}{ }^{\prime} p^{\alpha^{\prime}}, \tag{8.7}\\ \left\|L^{\beta}{ }_{\alpha^{\prime}}\right\|=\left\|L^{\alpha}{ }_{\beta}\right\|^{-1} ; \tag{8.8} \end{gather*}(8.7)eβ=eαLαβ,pβ=Lβαpα,(8.8)Lβα=Lαβ1;
Tensor algebra:
(1) occurs in infinitesimal neighborhood of an event
(2) is same in curved spacetime as in flat
(3) rules for component manipulation change slightly when using nonorthonormal basis
Components of metric
Transformation of basis
(4) in the special case of "coordinate bases," e α = P / x α , e β = P / x β e α = P / x α , e β = P / x β e_(alpha)=delP//delx^(alpha),e_(beta^('))=delP//delx^(beta^('))\boldsymbol{e}_{\alpha}=\partial \mathscr{P} / \partial x^{\alpha}, \boldsymbol{e}_{\beta^{\prime}}=\partial \mathscr{P} / \partial x^{\beta^{\prime}}eα=P/xα,eβ=P/xβ,
(8.9) L α β = x α / x β , L β α = x β / x α α ; (8.9) L α β = x α / x β , L β α = x β / x α α ; {:(8.9)L^(alpha^('))_(beta)=delx^(alpha^('))//delx^(beta)","quadL^(beta)_(alpha^('))=delx^(beta)//delx^(alpha^(alpha));:}\begin{equation*} L^{\alpha^{\prime}}{ }_{\beta}=\partial x^{\alpha^{\prime}} / \partial x^{\beta}, \quad L^{\beta}{ }_{\alpha^{\prime}}=\partial x^{\beta} / \partial x^{\alpha^{\alpha}} ; \tag{8.9} \end{equation*}(8.9)Lαβ=xα/xβ,Lβα=xβ/xαα;
Components of Levi-Civita tensor
and (5) the Levi-Civita tensor ε ε epsi\boldsymbol{\varepsilon}ε, like the metric tensor, has components that depend on how nonorthonormal the basis vectors are (see exercise 8.3): if e 0 e 0 e_(0)\boldsymbol{e}_{0}e0 points toward the future and e 1 , e 2 , e 3 e 1 , e 2 , e 3 e_(1),e_(2),e_(3)\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e1,e2,e3 are righthanded, then
(8.10a) ε α β γ δ = ( g ) 1 / 2 [ α β γ δ ] , ε α β γ δ = g 1 ε α β γ δ = ( g ) 1 / 2 [ α β γ δ ] , (8.10a) ε α β γ δ = ( g ) 1 / 2 [ α β γ δ ] , ε α β γ δ = g 1 ε α β γ δ = ( g ) 1 / 2 [ α β γ δ ] , {:[(8.10a)epsi_(alpha beta gamma delta)=(-g)^(1//2)[alpha beta gamma delta]","],[epsi^(alpha beta gamma delta)=g^(-1)epsi_(alpha beta gamma delta)=-(-g)^(-1//2)[alpha beta gamma delta]","]:}\begin{align*} \varepsilon_{\alpha \beta \gamma \delta} & =(-g)^{1 / 2}[\alpha \beta \gamma \delta], \tag{8.10a}\\ \varepsilon^{\alpha \beta \gamma \delta} & =g^{-1} \varepsilon_{\alpha \beta \gamma \delta}=-(-g)^{-1 / 2}[\alpha \beta \gamma \delta], \end{align*}(8.10a)εαβγδ=(g)1/2[αβγδ],εαβγδ=g1εαβγδ=(g)1/2[αβγδ],
where [ α β γ δ ] [ α β γ δ ] [alpha beta gamma delta][\alpha \beta \gamma \delta][αβγδ] is the completely antisymmetric symbol
(8.10b) [ α β γ δ ] { + 1 if α β γ δ is an even permutation of 0123, 1 if α β γ δ is an odd permutation of 0123 , 0 if α β γ δ are not all different, (8.10b) [ α β γ δ ] + 1  if  α β γ δ  is an even permutation of 0123,  1  if  α β γ δ  is an odd permutation of  0123 , 0  if  α β γ δ  are not all different,  {:(8.10b)[alpha beta gamma delta]-={[+1" if "alpha beta gamma delta" is an even permutation of 0123, "],[-1" if "alpha beta gamma delta" is an odd permutation of "0123","],[0" if "alpha beta gamma delta" are not all different, "]:}:}[\alpha \beta \gamma \delta] \equiv\left\{\begin{array}{c} +1 \text { if } \alpha \beta \gamma \delta \text { is an even permutation of 0123, } \tag{8.10b}\\ -1 \text { if } \alpha \beta \gamma \delta \text { is an odd permutation of } 0123, \\ 0 \text { if } \alpha \beta \gamma \delta \text { are not all different, } \end{array}\right.(8.10b)[αβγδ]{+1 if αβγδ is an even permutation of 0123, 1 if αβγδ is an odd permutation of 0123,0 if αβγδ are not all different, 
and where g g ggg is the determinant of the matrix g α β g α β ||g_(alpha beta)||\left\|g_{\alpha \beta}\right\|gαβ
(8.11) g det g α β = det e α e β . (8.11) g det g α β = det e α e β . {:(8.11)g-=det||g_(alpha beta)||=det||e_(alpha)*e_(beta)||.:}\begin{equation*} g \equiv \operatorname{det}\left\|g_{\alpha \beta}\right\|=\operatorname{det}\left\|\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}\right\| . \tag{8.11} \end{equation*}(8.11)gdetgαβ=deteαeβ.
Read Box 8.4 for full discussion and proofs; work exercise 8.1 below for fuller understanding and mastery.
Several dangers are glossed over in this discussion. In flat spacetime one often does not bother to say where a vector, 1 -form, or tensor is located. One freely moves geometric objects from event to event without even thinking. Of course, the unwritten rule of transport is: hold all lengths and directions fixed while moving; i.e., hold all Lorentz-frame components fixed; i.e., "parallel-transport" the object. But in

Box 8.4 TENSOR ALGEBRA AT A FIXED EVENT IN AN ARBITRARY BASIS

A. Bases

Tangent-vector basis: Pick e 0 , e 1 , e 2 , e 3 e 0 , e 1 , e 2 , e 3 e_(0),e_(1),e_(2),e_(3)\boldsymbol{e}_{0}, \boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e0,e1,e2,e3 at P 0 P 0 P_(0)\mathscr{P}_{0}P0 arbitrarily-but insist they be linearly independent.
"Dual basis" for 1-forms: The basis { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} determines a 1-form basis { ω α } ω α {omega^(alpha)}\left\{\boldsymbol{\omega}^{\alpha}\right\}{ωα} (its "dual basis") by
ω α , e β = δ α β ω α , e β = δ α β (:omega^(alpha),e_(beta):)=delta^(alpha)_(beta)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=\delta^{\alpha}{ }_{\beta}ωα,eβ=δαβ
[see equation (2.19)].
Geometric interpretation (Figure 9.2): e 2 , e 3 , e 0 e 2 , e 3 , e 0 e_(2),e_(3),e_(0)\boldsymbol{e}_{2}, \boldsymbol{e}_{3}, \boldsymbol{e}_{0}e2,e3,e0 lie parallel to surfaces of ω 1 ω 1 omega^(1)\boldsymbol{\omega}^{1}ω1; e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 pierces precisely one surface of ω 1 ω 1 omega^(1)\boldsymbol{\omega}^{1}ω1.
Function interpretation: ω α , e β = δ α β ω α , e β = δ α β (:omega^(alpha),e_(beta):)=delta^(alpha)_(beta)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=\delta^{\alpha}{ }_{\beta}ωα,eβ=δαβ determines the value of ω α ω α omega^(alpha)\boldsymbol{\omega}^{\alpha}ωα on any vector u = u β e β u = u β e β u=u^(beta)e_(beta)\boldsymbol{u}=u^{\beta} \boldsymbol{e}_{\beta}u=uβeβ (number of "bongs of bell" as u u u\boldsymbol{u}u pierces ω α ω α omega^(alpha)\boldsymbol{\omega}^{\alpha}ωα ):
ω α , u = ω α , u β e β = u β ω α , e β = u β δ α β = u α . ω α , u = ω α , u β e β = u β ω α , e β = u β δ α β = u α . (:omega^(alpha),u:)=(:omega^(alpha),u^(beta)e_(beta):)=u^(beta)(:omega^(alpha),e_(beta):)=u^(beta)delta^(alpha)_(beta)=u^(alpha).\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{u}\right\rangle=\left\langle\boldsymbol{\omega}^{\alpha}, u^{\beta} \boldsymbol{e}_{\beta}\right\rangle=u^{\beta}\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=u^{\beta} \delta^{\alpha}{ }_{\beta}=u^{\alpha} .ωα,u=ωα,uβeβ=uβωα,eβ=uβδαβ=uα.
Special case: coordinate bases. Choose an arbitrary coordinate system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)}. At P 0 P 0 P_(0)\mathscr{P}_{0}P0 choose e α = P / x α e α = P / x α e_(alpha)=delP//delx^(alpha)\boldsymbol{e}_{\alpha}=\partial \mathscr{P} / \partial x^{\alpha}eα=P/xα as basis vectors. Then the dual basis is ω α = d x α ω α = d x α omega^(alpha)=dx^(alpha)\boldsymbol{\omega}^{\alpha}=\boldsymbol{d} x^{\alpha}ωα=dxα. Proof: the general coordinate-free relation d f , v = v f d f , v = v f (:df,v:)=del_(v)f\langle\boldsymbol{d} f, \boldsymbol{v}\rangle=\partial_{\boldsymbol{v}} fdf,v=vf [equation (2.17)], with f = x α f = x α f=x^(alpha)f=x^{\alpha}f=xα and v = P / x β v = P / x β v=delP//delx^(beta)\boldsymbol{v}=\partial \mathscr{P} / \partial x^{\beta}v=P/xβ, reads
d x α , P / x β = ( / x β ) x α = δ α β . d x α , P / x β = / x β x α = δ α β . (:dx^(alpha),delP//delx^(beta):)=(del//delx^(beta))x^(alpha)=delta^(alpha)_(beta).\left\langle\boldsymbol{d} x^{\alpha}, \partial \mathscr{P} / \partial x^{\beta}\right\rangle=\left(\partial / \partial x^{\beta}\right) x^{\alpha}=\delta^{\alpha}{ }_{\beta} .dxα,P/xβ=(/xβ)xα=δαβ.

B. Algebra of Tangent Vectors and 1-Forms

The Lorentz-frame discussion of equations (2.19) to (2.22) is completely unchanged when one switches to an arbitrary basis. Its conclusions:
expansion, u = e α u α , σ = σ α ω α u = e α u α , σ = σ α ω α u=e_(alpha)u^(alpha),sigma=sigma_(alpha)omega^(alpha)\boldsymbol{u}=\boldsymbol{e}_{\alpha} u^{\alpha}, \boldsymbol{\sigma}=\sigma_{\alpha} \boldsymbol{\omega}^{\alpha}u=eαuα,σ=σαωα;
calculation of components, u α = ω α , u , σ α = σ , e α u α = ω α , u , σ α = σ , e α u^(alpha)=(:omega^(alpha),u:),sigma_(alpha)=(:sigma,e_(alpha):)u^{\alpha}=\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{u}\right\rangle, \sigma_{\alpha}=\left\langle\boldsymbol{\sigma}, \boldsymbol{e}_{\alpha}\right\rangleuα=ωα,u,σα=σ,eα;
value of form on vector, σ , u = σ α u α σ , u = σ α u α (:sigma,u:)=sigma_(alpha)u^(alpha)\langle\boldsymbol{\sigma}, \boldsymbol{u}\rangle=\sigma_{\alpha} u^{\alpha}σ,u=σαuα.
Application to gradients of functions:
expansion, d f = f , α ω α [ d f = f , α ω α df=f_(,alpha)omega^(alpha)[:}\boldsymbol{d} f=f_{, \alpha} \boldsymbol{\omega}^{\alpha}\left[\right.df=f,αωα[ defines f , α ] f , α {:f_(,alpha)]\left.f_{, \alpha}\right]f,α];
calculation of components, f , α = d f , e α = e α f f , α = d f , e α = e α f f_(,alpha)=(:df,e_(alpha):)=del_(e_(alpha))ff_{, \alpha}=\left\langle\boldsymbol{d} f, \boldsymbol{e}_{\alpha}\right\rangle=\partial_{\boldsymbol{e}_{\alpha}} ff,α=df,eα=eαf [see equation (2.17)].
Raising and lowering of indices is accomplished with g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ and g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ [equations (8.5) and (8.6)]. Proof:
u ~ u ~ widetilde(u)\widetilde{\boldsymbol{u}}u~, the 1 -form corresponding to u u u\boldsymbol{u}u, is defined by u ~ , v = u v u ~ , v = u v (: widetilde(u),v:)=u*v\langle\widetilde{\boldsymbol{u}}, \boldsymbol{v}\rangle=\boldsymbol{u} \cdot \boldsymbol{v}u~,v=uv for all v v v\boldsymbol{v}v; thus, u α u ~ , e α = u e α = u β e β e α = u β g β α ; u α u ~ , e α = u e α = u β e β e α = u β g β α ; u_(alpha)-=(:( widetilde(u)),e_(alpha):)=u*e_(alpha)=u^(beta)e_(beta)*e_(alpha)=u^(beta)g_(beta alpha);u_{\alpha} \equiv\left\langle\widetilde{\boldsymbol{u}}, \boldsymbol{e}_{\alpha}\right\rangle=\boldsymbol{u} \cdot \boldsymbol{e}_{\alpha}=u^{\beta} \boldsymbol{e}_{\beta} \cdot \boldsymbol{e}_{\alpha}=u^{\beta} g_{\beta \alpha} ;uαu~,eα=ueα=uβeβeα=uβgβα;
inverting this equation yields u β = g β α u α u β = g β α u α u^(beta)=g^(beta alpha)u_(alpha)u^{\beta}=g^{\beta \alpha} u_{\alpha}uβ=gβαuα.

C. Change of Basis

The discussion of Lorentz transformations in equations (2.39) to (2.43) is applicable to general changes of basis if one replaces Λ α β Λ α β ||Lambda^(alpha^('))_(beta)||\left\|\Lambda^{\alpha^{\prime}}{ }_{\beta}\right\|Λαβ by an arbitrary but nonsingular matrix L α β L α β ||L^(alpha^('))_(beta)||\left\|L^{\alpha^{\prime}}{ }_{\beta}\right\|Lαβ [equations (8.7), (8.8)]. Conclusions:
e α = e β L β α , e β = e α L α β ; ω α = L α β ω β , ω β = L β α ω α ; v α = L α α β v β , v β = L β α α v α ; σ α = σ β L β α , σ β = σ α L α β . e α = e β L β α , e β = e α L α β ; ω α = L α β ω β , ω β = L β α ω α ; v α = L α α β v β , v β = L β α α v α ; σ α = σ β L β α , σ β = σ α L α β . {:[e_(alpha^('))=e_(beta)L^(beta)_(alpha^('))","quade_(beta)=e_(alpha^('))L^(alpha^('))_(beta);],[omega^(alpha^('))=L^(alpha)_(beta)omega^(beta)","quadomega^(beta)=L^(beta)_(alpha)omega^(alpha^('));],[v^(alpha^('))=L^(alpha^(alpha))_(beta)v^(beta)","quadv^(beta)=L^(beta)_(alpha^(alpha))v^(alpha^('));],[sigma_(alpha^('))=sigma_(beta)L^(beta)_(alpha^('))","quadsigma_(beta)=sigma_(alpha^('))L^(alpha^('))_(beta).]:}\begin{aligned} & \boldsymbol{e}_{\alpha^{\prime}}=\boldsymbol{e}_{\beta} L^{\beta}{ }_{\alpha^{\prime}}, \quad \boldsymbol{e}_{\beta}=\boldsymbol{e}_{\alpha^{\prime}} L^{\alpha^{\prime}}{ }_{\beta} ; \\ & \boldsymbol{\omega}^{\alpha^{\prime}}=L^{\alpha}{ }_{\beta} \boldsymbol{\omega}^{\beta}, \quad \boldsymbol{\omega}^{\beta}=L^{\beta}{ }_{\alpha} \boldsymbol{\omega}^{\alpha^{\prime}} ; \\ & v^{\alpha^{\prime}}=L^{\alpha^{\alpha}}{ }_{\beta} v^{\beta}, \quad v^{\beta}=L^{\beta}{ }_{\alpha^{\alpha}} v^{\alpha^{\prime}} ; \\ & \sigma_{\alpha^{\prime}}=\sigma_{\beta} L^{\beta}{ }_{\alpha^{\prime}}, \quad \sigma_{\beta}=\sigma_{\alpha^{\prime}} L^{\alpha^{\prime}}{ }_{\beta} . \end{aligned}eα=eβLβα,eβ=eαLαβ;ωα=Lαβωβ,ωβ=Lβαωα;vα=Lααβvβ,vβ=Lβααvα;σα=σβLβα,σβ=σαLαβ.
When both bases are coordinate bases, then L β α = x β / x α , L α β = x α / x β L β α = x β / x α , L α β = x α / x β L^(beta)_(alpha^('))=delx^(beta)//delx^(alpha^(')),L^(alpha^('))_(beta)=delx^(alpha^('))//delx^(beta)L^{\beta}{ }_{\alpha^{\prime}}=\partial x^{\beta} / \partial x^{\alpha^{\prime}}, L^{\alpha^{\prime}}{ }_{\beta}=\partial x^{\alpha^{\prime}} / \partial x^{\beta}Lβα=xβ/xα,Lαβ=xα/xβ. Proof:
e α = x α = x β x α x β = x β x α e β ; similarly e β = x α x β e α . e α = x α = x β x α x β = x β x α e β ;  similarly  e β = x α x β e α . e_(alpha^('))=(del)/(delx^(alpha^(')))=(delx^(beta))/(delx^(alpha^(')))(del)/(delx^(beta))=(delx^(beta))/(delx^(alpha^(')))e_(beta);quad" similarly "e_(beta)=(delx^(alpha^(')))/(delx^(beta))e_(alpha^(')).\boldsymbol{e}_{\alpha^{\prime}}=\frac{\partial}{\partial x^{\alpha^{\prime}}}=\frac{\partial x^{\beta}}{\partial x^{\alpha^{\prime}}} \frac{\partial}{\partial x^{\beta}}=\frac{\partial x^{\beta}}{\partial x^{\alpha^{\prime}}} \boldsymbol{e}_{\beta} ; \quad \text { similarly } \boldsymbol{e}_{\beta}=\frac{\partial x^{\alpha^{\prime}}}{\partial x^{\beta}} \boldsymbol{e}_{\alpha^{\prime}} .eα=xα=xβxαxβ=xβxαeβ; similarly eβ=xαxβeα.

Box 8.4 (continued)

D. Algebra of Tensors

The discussions of tensor algebra given in § 3.2 § 3.2 §3.2\S 3.2§3.2 [equations (3.8) to (3.22)] and in § 3.5 § 3.5 §3.5\S 3.5§3.5 (excluding gradient and divergence) are unchanged, except that
η α β g α β η α β g α β , Λ α β L α β , Λ β α L β α ; η α β g α β η α β g α β , Λ α β L α β , Λ β α L β α ; eta_(alpha beta)longrightarrowg_(alpha beta)quadeta^(alpha beta)longrightarrowg^(alpha beta),quadLambda^(alpha^('))_(beta)longrightarrowL^(alpha^('))_(beta),quadLambda^(beta)_(alpha^('))longrightarrowL^(beta)_(alpha^('));\eta_{\alpha \beta} \longrightarrow g_{\alpha \beta} \quad \eta^{\alpha \beta} \longrightarrow g^{\alpha \beta}, \quad \Lambda^{\alpha^{\prime}}{ }_{\beta} \longrightarrow L^{\alpha^{\prime}}{ }_{\beta}, \quad \Lambda^{\beta}{ }_{\alpha^{\prime}} \longrightarrow L^{\beta}{ }_{\alpha^{\prime}} ;ηαβgαβηαβgαβ,ΛαβLαβ,ΛβαLβα;
and the components of the Levi-Civita tensor are changed from (3.50) to (8.10) [see exercise 8.3[.
Chief conclusions:
expansion, S = S α β γ e α ω β ω γ ; S = S α β γ e α ω β ω γ ; S=S^(alpha)_(beta gamma)e_(alpha)oxomega^(beta)oxomega^(gamma);\boldsymbol{S}=S^{\alpha}{ }_{\beta \gamma} \boldsymbol{e}_{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma} ;S=Sαβγeαωβωγ;
components, S α β γ = S ( ω α , e β , e γ ) S α β γ = S ω α , e β , e γ S^(alpha)_(beta gamma)=S(omega^(alpha),e_(beta),e_(gamma))S^{\alpha}{ }_{\beta \gamma}=\boldsymbol{S}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}\right)Sαβγ=S(ωα,eβ,eγ);
raising and lowering indices, S μ β ν = g μ α g ν γ S α β γ S μ β ν = g μ α g ν γ S α β γ S_(mu beta)^(nu)=g_(mu alpha)g^(nu gamma)S^(alpha)_(beta gamma)S_{\mu \beta}{ }^{\nu}=g_{\mu \alpha} g^{\nu \gamma} S^{\alpha}{ }_{\beta \gamma}Sμβν=gμαgνγSαβγ;
change of basis, S λ μ ν = L λ α L β μ L γ ν S α β γ S λ μ ν = L λ α L β μ L γ ν S α β γ S^(lambda^('))_(mu^(')nu^('))=L^(lambda^('))_(alpha)L^(beta)_(mu^('))L^(gamma)_(nu)^(')S^(alpha)_(beta gamma)S^{\lambda^{\prime}}{ }_{\mu^{\prime} \nu^{\prime}}=L^{\lambda^{\prime}}{ }_{\alpha} L^{\beta}{ }_{\mu^{\prime}} L^{\gamma}{ }_{\nu}{ }^{\prime} S^{\alpha}{ }_{\beta \gamma}Sλμν=LλαLβμLγνSαβγ;
machine operation, S ( σ , u , v ) = S α β γ σ α u β v γ ; S ( σ , u , v ) = S α β γ σ α u β v γ ; S(sigma,u,v)=S^(alpha)_(beta gamma)^(sigma)_(alpha)u^(beta)v^(gamma);\boldsymbol{S}(\sigma, \boldsymbol{u}, \boldsymbol{v})=S^{\alpha}{ }_{\beta \gamma}{ }^{\sigma}{ }_{\alpha} u^{\beta} v^{\gamma} ;S(σ,u,v)=Sαβγσαuβvγ;
tensor product, T = u v T α β = u α v β ; T = u v T α β = u α v β ; T=u ox v LongleftrightarrowT^(alpha beta)=u^(alpha)v^(beta);\boldsymbol{T}=\boldsymbol{u} \otimes \boldsymbol{v} \Longleftrightarrow T^{\alpha \beta}=u^{\alpha} v^{\beta} ;T=uvTαβ=uαvβ;
contraction, " M = M = M=\boldsymbol{M}=M= contraction of R R R\boldsymbol{R}R on slots 1 and 3 " M μ ν = R α μ α ν 3 " M μ ν = R α μ α ν 3"LongleftrightarrowM_(mu nu)=R^(alpha)_(mu alpha nu)3 " \Longleftrightarrow M_{\mu \nu}=R^{\alpha}{ }_{\mu \alpha \nu}3"Mμν=Rαμαν;
wedge product, α β α β alpha^^beta\boldsymbol{\alpha} \wedge \boldsymbol{\beta}αβ has components α μ β ν β μ α ν α μ β ν β μ α ν alpha^(mu)beta^(nu)-beta^(mu)alpha^(nu)\alpha^{\mu} \beta^{\nu}-\beta^{\mu} \alpha^{\nu}αμβνβμαν;
Dual, J α β γ = J μ ε μ α β γ , F α β = 1 2 F μ ν ε μ ν α β , B α = 1 6 B λ μ ν ε λ μ ν α J α β γ = J μ ε μ α β γ , F α β = 1 2 F μ ν ε μ ν α β , B α = 1 6 B λ μ ν ε λ μ ν α ^(**)J_(alpha beta gamma)=J^(mu)epsi_(mu alpha beta gamma),^(**)F_(alpha beta)=(1)/(2)F^(mu nu)epsi_(mu nu alpha beta),^(**)B_(alpha)=(1)/(6)B^(lambda mu nu)epsi_(lambda mu nu alpha){ }^{*} J_{\alpha \beta \gamma}=J^{\mu} \varepsilon_{\mu \alpha \beta \gamma},{ }^{*} F_{\alpha \beta}=\frac{1}{2} F^{\mu \nu} \varepsilon_{\mu \nu \alpha \beta},{ }^{*} B_{\alpha}=\frac{1}{6} B^{\lambda \mu \nu} \varepsilon_{\lambda \mu \nu \alpha}Jαβγ=Jμεμαβγ,Fαβ=12Fμνεμναβ,Bα=16Bλμνελμνα.

E. Commutators (exercise 8.2; §9.6; Box 9.2)

If u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are tangent vector fields, one takes the view that u = u u = u u=del_(u)\boldsymbol{u}=\partial_{\boldsymbol{u}}u=u and v = v v = v v=del_(v)\boldsymbol{v}=\partial_{\boldsymbol{v}}v=v, and one defines
[ u , v ] [ u , v ] u v v u [ u , v ] u , v u v v u [u,v]-=[del_(u),del_(v)]-=del_(u)del_(v)-del_(v)del_(u)[\boldsymbol{u}, \boldsymbol{v}] \equiv\left[\partial_{\boldsymbol{u}}, \partial_{\mathbf{v}}\right] \equiv \partial_{\boldsymbol{u}} \partial_{\boldsymbol{v}}-\partial_{\boldsymbol{v}} \partial_{\boldsymbol{u}}[u,v][u,v]uvvu
This commutator is itself a tangent vector field.
Components in a coordinate basis:
[ u , v ] = ( u β v α , β v β u α , β ) ( / x α ) . [ = e α ] [ u , v ] = u β v α , β v β u α , β / x α . = e α {:[[u","v]=(u^(beta)v^(alpha)_(,beta)-v^(beta)u^(alpha)_(,beta))(del//delx^(alpha)).],[llcorner[=e_(alpha)]:}]:}\begin{array}{r} {[\boldsymbol{u}, \boldsymbol{v}]=\left(u^{\beta} v^{\alpha}{ }_{, \beta}-v^{\beta} u^{\alpha}{ }_{, \beta}\right)\left(\partial / \partial x^{\alpha}\right) .} \\ \left\llcorner\left[=\boldsymbol{e}_{\alpha}\right]\right. \end{array}[u,v]=(uβvα,βvβuα,β)(/xα).[=eα]
Commutation coefficients of a basis:
[ e α , e β ] c α β γ e γ , c α β μ c α β γ g γ μ e α , e β c α β γ e γ , c α β μ c α β γ g γ μ [e_(alpha),e_(beta)]-=c_(alpha beta)^(gamma)e_(gamma),quadc_(alpha beta mu)-=c_(alpha beta)^(gamma)g_(gamma mu)\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right] \equiv c_{\alpha \beta}{ }^{\gamma} \boldsymbol{e}_{\gamma}, \quad c_{\alpha \beta \mu} \equiv c_{\alpha \beta}{ }^{\gamma} g_{\gamma \mu}[eα,eβ]cαβγeγ,cαβμcαβγgγμ
Coordinate basis ("holonomic") c α β γ = 0 c α β γ = 0 c_(alpha beta)^(gamma)=0c_{\alpha \beta}{ }^{\gamma}=0cαβγ=0;
Noncoordinate basis ("anholonomic") some c α β γ 0 c α β γ 0 c_(alpha beta)^(gamma)!=0c_{\alpha \beta}{ }^{\gamma} \neq 0cαβγ0 (see exercise 9.9).
curved spacetime there is no global Lorentz coordinate system in which to hold components fixed; and objects initially parallel, after "parallel transport" along different curves cease to be parallel ("geodesic deviation"; Earth's meridians, parallel at equator, cross at north and south poles). Thus, in curved spacetime one must not blithely move a geometric object from point to point, without carefully specifying how it is to be moved and by what route. Each local geometric object has its own official place of residence (event P o P o P_(o)\mathscr{P}_{o}Po ); it can interact with other objects residing there (tensor algebra); but it cannot interact with any object at another event Q Q Q\mathcal{Q}Q, until it has been carefully transported from P o P o P_(o)\mathscr{P}_{o}Po to Q Q Q\mathcal{Q}Q.
This line of reasoning, pursued further, leads one to speak of the "tangent space" at each event, in which that event's vectors (arrows) and 1 -forms (families of surfaces) lie, and in which its tensors (linear machines) operate. One even draws heuristic pictures of the tangent space, as in Figure 9.1 (p. 231).
Another danger in curved spacetime is the temptation to regard vectors as arrows linking two events ("point for head and point for tail")-i.e., to regard the tangent space of Figure 9.1 as lying in spacetime itself. This practice can be useful for heuristic purposes, but it is incompatible with complete mathematical precision. (How is the tangent space to be molded into a warped surface?) Four definitions of a vector were given in Figure 2.1 (page 49): three definitions relying on "point for head and point for tail"; one, " d P / d λ " d P / d λ " dP//d lambda"d \mathscr{P} / d \lambda "dP/dλ" ", purely local. Only the local definition is wholly viable in curved spacetime, and even it can be improved upon, in the eyes of mathematicians, as follows.
There is a one-to-one correspondence (complete "isomorphism") between vectors u u u\boldsymbol{u}u and directional derivative operators u u del_(u)\partial_{\boldsymbol{u}}u. The concept of vector is a bit fuzzy, but "directional derivative" is perfectly well-defined. To get rid of all fuzziness, exploit the isomorphism to the full: define the tangent vector u u u\boldsymbol{u}u to be equal to the corresponding directional derivative
(8.12) u u (8.12) u u {:(8.12)u-=del_(u):}\begin{equation*} \boldsymbol{u} \equiv \partial_{\boldsymbol{u}} \tag{8.12} \end{equation*}(8.12)uu
(This practice, unfamiliar as it may be to a physicist at first, has mathematical power; so this book will use it frequently. For a fuller discussion, see §9.2.)

Exercise 8.1. PRACTICE WITH TENSOR ALGEBRA

EXERCISES

Let t , x , y , z t , x , y , z t,x,y,zt, x, y, zt,x,y,z be Lorentz coordinates in flat spacetime, and let
r = ( x 2 + y 2 + z 2 ) 1 / 2 , θ = cos 1 ( z / r ) , ϕ = tan 1 ( y / x ) r = x 2 + y 2 + z 2 1 / 2 , θ = cos 1 ( z / r ) , ϕ = tan 1 ( y / x ) r=(x^(2)+y^(2)+z^(2))^(1//2),quad theta=cos^(-1)(z//r),quad phi=tan^(-1)(y//x)r=\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}, \quad \theta=\cos ^{-1}(z / r), \quad \phi=\tan ^{-1}(y / x)r=(x2+y2+z2)1/2,θ=cos1(z/r),ϕ=tan1(y/x)
be the corresponding spherical coordinates. Then
e 0 = P / t , e r = P / r , e θ = P / θ , e ϕ = P / ϕ e 0 = P / t , e r = P / r , e θ = P / θ , e ϕ = P / ϕ e_(0)=delP//del t,quade_(r)=delP//del r,quade_(theta)=delP//del theta,quade_(phi)=delP//del phi\boldsymbol{e}_{0}=\partial \mathscr{P} / \partial t, \quad \boldsymbol{e}_{r}=\partial \mathscr{P} / \partial r, \quad \boldsymbol{e}_{\theta}=\partial \mathscr{P} / \partial \theta, \quad \boldsymbol{e}_{\phi}=\partial \mathscr{P} / \partial \phie0=P/t,er=P/r,eθ=P/θ,eϕ=P/ϕ
is a coordinate basis, and
e 0 ^ = P t , e r ˙ = P r , e θ ` = 1 r P θ , e ϕ ^ = 1 r sin θ P ϕ e 0 ^ = P t , e r ˙ = P r , e θ ` = 1 r P θ , e ϕ ^ = 1 r sin θ P ϕ e_( hat(0))=(delP)/(del t),quade_(r^(˙))=(delP)/(del r),quade_(theta^(`))=(1)/(r)(delP)/(del theta),quade_( hat(phi))=(1)/(r sin theta)(delP)/(del phi)\boldsymbol{e}_{\hat{0}}=\frac{\partial \mathscr{P}}{\partial t}, \quad \boldsymbol{e}_{\dot{r}}=\frac{\partial \mathscr{P}}{\partial r}, \quad \boldsymbol{e}_{\grave{\theta}}=\frac{1}{r} \frac{\partial \mathscr{P}}{\partial \theta}, \quad \boldsymbol{e}_{\hat{\phi}}=\frac{1}{r \sin \theta} \frac{\partial \mathscr{P}}{\partial \phi}e0^=Pt,er˙=Pr,eθ`=1rPθ,eϕ^=1rsinθPϕ
is a noncoordinate basis.
Vectors and tensors must not be moved blithely from point to point
Tangent space defined
Definitions of vector in curved spacetime:
(1) as d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ
(2) as directional derivative
(a) Draw a picture of e θ , e ϕ , e θ ^ e θ , e ϕ , e θ ^ e_(theta),e_(phi),e_( hat(theta))\boldsymbol{e}_{\theta}, \boldsymbol{e}_{\phi}, \boldsymbol{e}_{\hat{\theta}}eθ,eϕ,eθ^, and e ϕ ^ e ϕ ^ e_( hat(phi))\boldsymbol{e}_{\hat{\phi}}eϕ^ at several different points on a sphere of constant t , r t , r t,rt, rt,r. [Answer for e θ , e ϕ e θ , e ϕ e_(theta),e_(phi)\boldsymbol{e}_{\theta}, \boldsymbol{e}_{\phi}eθ,eϕ should resemble Figure 9.1.]
(b) What are the 1-form bases { ω α } ω α {omega^(alpha)}\left\{\boldsymbol{\omega}^{\alpha}\right\}{ωα} and { ω α ^ } ω α ^ {omega^( hat(alpha))}\left\{\boldsymbol{\omega}^{\hat{\alpha}}\right\}{ωα^} dual to these tangent-vector bases? [Answer: ω 0 = d t , ω r = d r , ω θ = d θ , ω ϕ = d ϕ ; ω 0 ^ = d t , ω i ^ = d r , ω θ ^ = r d θ , ω ϕ ^ = r sin θ d ϕ . ] ω 0 = d t , ω r = d r , ω θ = d θ , ω ϕ = d ϕ ; ω 0 ^ = d t , ω i ^ = d r , ω θ ^ = r d θ , ω ϕ ^ = r sin θ d ϕ . {:omega^(0)=dt,omega^(r)=dr,omega^(theta)=d theta,omega^(phi)=d phi;omega^( hat(0))=dt,omega^( hat(i))=dr,omega^( hat(theta))=rd theta,omega^( hat(phi))=r sin theta d phi.]\left.\boldsymbol{\omega}^{0}=\boldsymbol{d} t, \boldsymbol{\omega}^{r}=\boldsymbol{d} r, \boldsymbol{\omega}^{\theta}=\boldsymbol{d} \theta, \boldsymbol{\omega}^{\phi}=\boldsymbol{d} \phi ; \boldsymbol{\omega}^{\hat{0}}=\boldsymbol{d} t, \boldsymbol{\omega}^{\hat{i}}=\boldsymbol{d} r, \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} \theta, \boldsymbol{\omega}^{\hat{\phi}}=r \sin \theta \boldsymbol{d} \phi.\right]ω0=dt,ωr=dr,ωθ=dθ,ωϕ=dϕ;ω0^=dt,ωi^=dr,ωθ^=rdθ,ωϕ^=rsinθdϕ.]
(c) What is the transformation matrix linking the original Lorentz frame to the spherical coordinate frame { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} ? [Answer: nonzero components are
L 0 t = 1 , L r z = z r = cos θ , L θ z = z θ = r sin θ , L r x = sin θ cos ϕ , L θ x = r cos θ cos ϕ , L ϕ x = r sin θ sin ϕ L y r = sin θ sin ϕ , L θ y = r cos θ sin ϕ , L ϕ y = r sin θ cos ϕ . ] L 0 t = 1 ,      L r z = z r = cos θ ,      L θ z = z θ = r sin θ , L r x = sin θ cos ϕ ,      L θ x = r cos θ cos ϕ ,      L ϕ x = r sin θ sin ϕ L y r = sin θ sin ϕ ,      L θ y = r cos θ sin ϕ ,      L ϕ y = r sin θ cos ϕ . {:[L_(0)^(t)=1",",L_(r)^(z)=(del z)/(del r)=cos theta",",L_(theta)^(z)=(del z)/(del theta)=-r sin theta","],[L_(r)^(x)=sin theta cos phi",",L_(theta)^(x)=r cos theta cos phi",",L_(phi)^(x)=-r sin theta sin phi],[L^(y)_(r)=sin theta sin phi",",L_(theta)^(y)=r cos theta sin phi",",{:L_(phi)^(y)=r sin theta cos phi.]]:}\begin{array}{lll} L_{0}^{t}=1, & L_{r}^{z}=\frac{\partial z}{\partial r}=\cos \theta, & L_{\theta}^{z}=\frac{\partial z}{\partial \theta}=-r \sin \theta, \\ L_{r}^{x}=\sin \theta \cos \phi, & L_{\theta}^{x}=r \cos \theta \cos \phi, & L_{\phi}^{x}=-r \sin \theta \sin \phi \\ L^{y}{ }_{r}=\sin \theta \sin \phi, & L_{\theta}^{y}=r \cos \theta \sin \phi, & \left.L_{\phi}^{y}=r \sin \theta \cos \phi .\right] \end{array}L0t=1,Lrz=zr=cosθ,Lθz=zθ=rsinθ,Lrx=sinθcosϕ,Lθx=rcosθcosϕ,Lϕx=rsinθsinϕLyr=sinθsinϕ,Lθy=rcosθsinϕ,Lϕy=rsinθcosϕ.]
(d) Use this transformation matrix to calculate the metric components g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ in the spherical coordinate basis, and invert the resulting matrix to get g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ. [Answer:
g 00 = 1 , g r r = 1 , g θ θ = r 2 , g ϕ ϕ = r 2 sin 2 θ , all other g α β = 0 g 00 = 1 ,      g r r = 1 ,      g θ θ = r 2 ,      g ϕ ϕ = r 2 sin 2 θ ,  all other  g α β = 0 {:[g_(00)=-1",",g_(rr)=1",",g_(theta theta)=r^(2)",",g_(phi phi)=r^(2)sin^(2)theta","]:}quad" all other "g_(alpha beta)=0\begin{array}{llll} g_{00}=-1, & g_{r r}=1, & g_{\theta \theta}=r^{2}, & g_{\phi \phi}=r^{2} \sin ^{2} \theta, \end{array} \quad \text { all other } g_{\alpha \beta}=0g00=1,grr=1,gθθ=r2,gϕϕ=r2sin2θ, all other gαβ=0
(e) Show that the noncoordinate basis { e α ^ } e α ^ {e_( hat(alpha))}\left\{\boldsymbol{e}_{\hat{\alpha}}\right\}{eα^} is orthonormal everywhere; i.e., that g α ^ β ^ = g α ^ β ^ = g_( hat(alpha) hat(beta))=g_{\hat{\alpha} \hat{\beta}}=gα^β^= η α β η α β eta_(alpha beta)\eta_{\alpha \beta}ηαβ; i.e. that
g = ω 0 ^ ω ^ ^ + ω ı ^ ω ı ^ + ω θ ^ ω t ^ ^ + ω ϕ ^ ω ^ ^ . g = ω 0 ^ ω ^ ^ + ω ı ^ ω ı ^ + ω θ ^ ω t ^ ^ + ω ϕ ^ ω ^ ^ . g=-omega^( hat(0))oxomega^( hat(hat()))+omega^( hat(ı))oxomega^( hat(ı))+omega^( hat(theta))oxomega^( hat(hat(t)))+omega^( hat(phi))oxomega^( hat(hat())).\boldsymbol{g}=-\boldsymbol{\omega}^{\hat{0}} \otimes \boldsymbol{\omega}^{\hat{\hat{}}}+\boldsymbol{\omega}^{\hat{\imath}} \otimes \boldsymbol{\omega}^{\hat{\imath}}+\boldsymbol{\omega}^{\hat{\theta}} \otimes \boldsymbol{\omega}^{\hat{\hat{t}}}+\boldsymbol{\omega}^{\hat{\phi}} \otimes \boldsymbol{\omega}^{\hat{\hat{}}} .g=ω0^ω^^+ωı^ωı^+ωθ^ωt^^+ωϕ^ω^^.
(f) Write the gradient of a function f f fff in terms of the spherical coordinate and noncoordinate bases. [Answer:
d f = f t d t + f r d r + f θ d θ + f ϕ d ϕ = f t ω 0 ^ + f r ω r ^ + 1 r f θ ω θ ^ + 1 r sin θ f ϕ ω ϕ ^ . ] d f = f t d t + f r d r + f θ d θ + f ϕ d ϕ = f t ω 0 ^ + f r ω r ^ + 1 r f θ ω θ ^ + 1 r sin θ f ϕ ω ϕ ^ . {:[df=(del f)/(del t)dt+(del f)/(del r)dr+(del f)/(del theta)d theta+(del f)/(del phi)d phi],[{:=(del f)/(del t)omega^( hat(0))+(del f)/(del r)omega^( hat(r))+(1)/(r)(del f)/(del theta)omega^( hat(theta))+(1)/(r sin theta)(del f)/(del phi)omega^( hat(phi)).]]:}\begin{aligned} \boldsymbol{d} f & =\frac{\partial f}{\partial t} \boldsymbol{d} t+\frac{\partial f}{\partial r} \boldsymbol{d} r+\frac{\partial f}{\partial \theta} \boldsymbol{d} \theta+\frac{\partial f}{\partial \phi} \boldsymbol{d} \phi \\ & \left.=\frac{\partial f}{\partial t} \boldsymbol{\omega}^{\hat{0}}+\frac{\partial f}{\partial r} \boldsymbol{\omega}^{\hat{r}}+\frac{1}{r} \frac{\partial f}{\partial \theta} \boldsymbol{\omega}^{\hat{\theta}}+\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \boldsymbol{\omega}^{\hat{\phi}} .\right] \end{aligned}df=ftdt+frdr+fθdθ+fϕdϕ=ftω0^+frωr^+1rfθωθ^+1rsinθfϕωϕ^.]
(g) What are the components of the Levi-Civita tensor in the spherical coordinate and noncoordinate bases? [Answer for coordinate basis:
ε 0 r θ ϕ = ε r 0 θ ϕ = + ε r θ 0 ϕ = = r 2 sin θ ε θ r θ ϕ = ε r 0 θ ϕ = + ε r θ ϕ 0 = = r 2 sin 1 θ . ] ε 0 r θ ϕ = ε r 0 θ ϕ = + ε r θ 0 ϕ = = r 2 sin θ ε θ r θ ϕ = ε r 0 θ ϕ = + ε r θ ϕ 0 = = r 2 sin 1 θ . {:[epsi_(0r theta phi)=-epsi_(r0theta phi)=+epsi_(r theta0phi)=cdots=r^(2)sin theta],[{:epsi^(theta r theta phi)=-epsi^(r0theta phi)=+epsi^(r theta phi0)=cdots=-r^(-2)sin^(-1)theta.]]:}\begin{aligned} & \varepsilon_{0 r \theta \phi}=-\varepsilon_{r 0 \theta \phi}=+\varepsilon_{r \theta 0 \phi}=\cdots=r^{2} \sin \theta \\ & \left.\varepsilon^{\theta r \theta \phi}=-\varepsilon^{r 0 \theta \phi}=+\varepsilon^{r \theta \phi 0}=\cdots=-r^{-2} \sin ^{-1} \theta .\right] \end{aligned}ε0rθϕ=εr0θϕ=+εrθ0ϕ==r2sinθεθrθϕ=εr0θϕ=+εrθϕ0==r2sin1θ.]

Exercise 8.2. COMMUTATORS

Take the mathematician's viewpoint that tangent vectors and directional derivatives are the same thing, u u u u u-=del_(u)\boldsymbol{u} \equiv \partial_{\boldsymbol{u}}uu. Let u u u\boldsymbol{u}u and v v v\boldsymbol{v}v be two vector fields, and define their commutator in the manner familiar from quantum mechanics
(8.13a) [ u , v ] [ u , v ] u v v u (8.13a) [ u , v ] u , v u v v u {:(8.13a)[u","v]-=[del_(u),del_(v)]-=del_(u)del_(v)-del_(v)del_(u):}\begin{equation*} [\boldsymbol{u}, \boldsymbol{v}] \equiv\left[\partial_{\boldsymbol{u}}, \partial_{\boldsymbol{v}}\right] \equiv \partial_{\boldsymbol{u}} \partial_{\boldsymbol{v}}-\partial_{\boldsymbol{v}} \partial_{\boldsymbol{u}} \tag{8.13a} \end{equation*}(8.13a)[u,v][u,v]uvvu
(a) Derive the following expression for [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v], valid in any coordinate basis,
(8.13b) [ u , v ] = ( u β v α , β v β u α , β ) e α (8.13b) [ u , v ] = u β v α , β v β u α , β e α {:(8.13b)[u","v]=(u^(beta)v^(alpha)_(,beta)-v^(beta)u^(alpha)_(,beta))e_(alpha):}\begin{equation*} [\boldsymbol{u}, \boldsymbol{v}]=\left(u^{\beta} v^{\alpha}{ }_{, \beta}-v^{\beta} u^{\alpha}{ }_{, \beta}\right) \boldsymbol{e}_{\alpha} \tag{8.13b} \end{equation*}(8.13b)[u,v]=(uβvα,βvβuα,β)eα
Thus, despite the fact that it looks like a second-order differential operator, [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] is actually of first order-i.e., it is a tangent vector.
(b) For any basis { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα}, one defines the "commutation coefficients" c β γ α c β γ α c_(beta gamma)^(alpha)c_{\beta \gamma}{ }^{\alpha}cβγα and c β γ α c β γ α c_(beta gamma alpha)c_{\beta \gamma \alpha}cβγα by
(8.14) [ e β , e γ ] c β γ α e α ; c β γ α g α μ c β γ μ (8.14) e β , e γ c β γ α e α ; c β γ α g α μ c β γ μ {:(8.14)[e_(beta),e_(gamma)]-=c_(beta gamma)^(alpha)e_(alpha);quadc_(beta gamma alpha)-=g_(alpha mu)c_(beta gamma)^(mu):}\begin{equation*} \left[\boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}\right] \equiv c_{\beta \gamma}{ }^{\alpha} \boldsymbol{e}_{\alpha} ; \quad c_{\beta \gamma \alpha} \equiv g_{\alpha \mu} c_{\beta \gamma}{ }^{\mu} \tag{8.14} \end{equation*}(8.14)[eβ,eγ]cβγαeα;cβγαgαμcβγμ
Show that c β γ α = c β γ α = 0 c β γ α = c β γ α = 0 c_(beta gamma)^(alpha)=c_(beta gamma alpha)=0c_{\beta \gamma}{ }^{\alpha}=c_{\beta \gamma \alpha}=0cβγα=cβγα=0 for any coordinate basis.
(c) Calculate c β ^ γ ^ α ^ c β ^ γ ^ α ^ c_( hat(beta) hat(gamma))^( hat(alpha))c_{\hat{\beta} \hat{\gamma}}{ }^{\hat{\alpha}}cβ^γ^α^ for the spherical noncoordinate basis of exercise 8.1. [Answer: All vanish except
c r ^ θ ^ θ ^ = c θ ^ r ^ θ ^ = 1 / r c r ^ ϕ ^ ϕ ^ = c ϕ ^ r ^ ϕ ^ = 1 / r c θ ^ ϕ ^ ϕ ^ = c ϕ ^ θ ^ θ ^ β ^ = cot θ / r . ] c r ^ θ ^ θ ^ = c θ ^ r ^ θ ^ = 1 / r c r ^ ϕ ^ ϕ ^ = c ϕ ^ r ^ ϕ ^ = 1 / r c θ ^ ϕ ^ ϕ ^ = c ϕ ^ θ ^ θ ^ β ^ = cot θ / r . {:[c_( hat(r) hat(theta))^( hat(theta))=-c_( hat(theta) hat(r))^( hat(theta))=-1//r],[c_( hat(r) hat(phi)) hat(phi)=-c_( hat(phi) hat(r))^( hat(phi))=-1//r],[c_( hat(theta) hat(phi))^( hat(phi)){:=-c_( hat(phi) hat(theta)) hat(theta)^( hat(beta))=-cot theta//r.]]:}\begin{aligned} c_{\hat{r} \hat{\theta}}^{\hat{\theta}} & =-c_{\hat{\theta} \hat{r}}^{\hat{\theta}}=-1 / r \\ c_{\hat{r} \hat{\phi}} \hat{\phi} & =-c_{\hat{\phi} \hat{r}}^{\hat{\phi}}=-1 / r \\ c_{\hat{\theta} \hat{\phi}}^{\hat{\phi}} & \left.=-c_{\hat{\phi} \hat{\theta}} \hat{\theta}^{\hat{\beta}}=-\cot \theta / r .\right] \end{aligned}cr^θ^θ^=cθ^r^θ^=1/rcr^ϕ^ϕ^=cϕ^r^ϕ^=1/rcθ^ϕ^ϕ^=cϕ^θ^θ^β^=cotθ/r.]

Exercise 8.3. COMPONENTS OF LEVI-CIVITA TENSOR IN NONORTHONORMAL FRAME

(a) Show that expressions (8.10) are the components of ε ε epsi\boldsymbol{\varepsilon}ε in an arbitrary basis, with e 0 e 0 e_(0)\boldsymbol{e}_{0}e0 pointing toward the future and e 1 , e 2 , e 3 e 1 , e 2 , e 3 e_(1),e_(2),e_(3)\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e1,e2,e3 right-handed. [Hints: (1) Review the discussion of ε ε epsi\varepsilonε in Lorentz frames, given in exercise 3.13. (2) Calculate ε α β γ δ ε α β γ δ epsi_(alpha beta gamma delta)\varepsilon_{\alpha \beta \gamma \delta}εαβγδ and ε α β γ δ ε α β γ δ epsi^(alpha beta gamma delta)\varepsilon^{\alpha \beta \gamma \delta}εαβγδ by transforming from a local Lorentz frame { e μ ^ } e μ ^ {e_( hat(mu))}\left\{\boldsymbol{e}_{\hat{\mu}}\right\}{eμ^}, e.g.,
ε α β γ δ = L α μ ^ L β p ^ L L ^ γ L ρ ^ δ ε μ ^ ν ^ ρ ^ ρ ^ . ε α β γ δ = L α μ ^ L β p ^ L L ^ γ L ρ ^ δ ε μ ^ ν ^ ρ ^ ρ ^ . epsi_(alpha beta gamma delta)=L_(alpha)^( hat(mu))L_(beta)^( hat(p))L^( hat(L))_(gamma)L^( hat(rho))_(delta)epsi_( hat(mu) hat(nu) hat(rho) hat(rho)).\varepsilon_{\alpha \beta \gamma \delta}=L_{\alpha}^{\hat{\mu}} L_{\beta}^{\hat{p}} L^{\hat{L}}{ }_{\gamma} L^{\hat{\rho}}{ }_{\delta} \varepsilon_{\hat{\mu} \hat{\nu} \hat{\rho} \hat{\rho}} .εαβγδ=Lαμ^Lβp^LL^γLρ^δεμ^ν^ρ^ρ^.
(3) Show that these expressions reduce to
ε α β γ δ = det L μ ^ p ε α ^ β ^ γ ^ δ , ε α β γ δ = det L ν μ ^ μ ^ ε α ^ β ^ γ ^ . ε α β γ δ = det L μ ^ p ε α ^ β ^ γ ^ δ , ε α β γ δ = det L ν μ ^ μ ^ ε α ^ β ^ γ ^ . epsi_(alpha beta gamma delta)=det||L^( hat(mu)_(p))||epsi_( hat(alpha) hat(beta) hat(gamma)delta),quadepsi^(alpha beta gamma delta)=det||L^(nu)|| hat(mu)^( hat(mu))epsi^( hat(alpha) hat(beta) hat(gamma)).\varepsilon_{\alpha \beta \gamma \delta}=\operatorname{det}\left\|L^{\hat{\mu}_{p}}\right\| \varepsilon_{\hat{\alpha} \hat{\beta} \hat{\gamma} \delta}, \quad \varepsilon^{\alpha \beta \gamma \delta}=\operatorname{det}\left\|L^{\nu}\right\| \hat{\mu}^{\hat{\mu}} \varepsilon^{\hat{\alpha} \hat{\beta} \hat{\gamma}} .εαβγδ=detLμ^pεα^β^γ^δ,εαβγδ=detLνμ^μ^εα^β^γ^.
(4) Show, from the transformation law for the metric components, that
( det L μ ^ μ ^ ) 2 det g α β = 1 det L μ ^ μ ^ 2 det g α β = 1 (det||L_( hat(mu))_( hat(mu))||)^(2)det||g_(alpha beta)||=-1\left(\operatorname{det}\left\|L_{\hat{\mu}}{ }_{\hat{\mu}}\right\|\right)^{2} \operatorname{det}\left\|g_{\alpha \beta}\right\|=-1(detLμ^μ^)2detgαβ=1
(5) Combine these results to obtain expressions (8.10).]
(b) Show that the components of the permutation tensors [defined by equations ( 3.50 h ) ( 3.50 h ) (3.50h)-(3.50 \mathrm{~h})-(3.50 h) ( 3.50 j ) ( 3.50 j ) (3.50j)(3.50 \mathrm{j})(3.50j) ] have the same values [equations ( 3.50 k ) ( 3.50 m ) ( 3.50 k ) ( 3.50 m ) (3.50k)-(3.50m)(3.50 \mathrm{k})-(3.50 \mathrm{~m})(3.50k)(3.50 m) ] in arbitrary frames as in Lorentz frames.
Additional exercises on tensor algebra: exercises 9.3 and 9.4 (page 234).

§8.5. PARALLEL TRANSPORT, COVARIANT DERIVATIVE, CONNECTION COEFFICIENTS, GEODESICS

The vehicle that carries one from classical mechanics to quantum mechanics is the correspondence principle. Similarly, the vehicle between flat spacetime and curved spacetime is the equivalence principle: "The laws of physics are the same in any local Lorentz frame of curved spacetime as in a global Lorentz frame of flat spacetime." But to apply the equivalence principle, one must first have a mathematical representation of a local Lorentz frame. The obvious choice is this: A local Lorentz frame at a given event P o P o P_(o)\mathscr{P}_{o}Po is the closest thing there is to a global Lorentz frame at that event; i.e., it is a coordinate system in which
(8.15a) g μ v ( P o ) = η μ v (8.15a) g μ v P o = η μ v {:(8.15a)g_(mu v)(P_(o))=eta_(mu v):}\begin{equation*} g_{\mu v}\left(\mathscr{P}_{o}\right)=\eta_{\mu v} \tag{8.15a} \end{equation*}(8.15a)gμv(Po)=ημv
and in which g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν holds as tightly as possible to η μ ν η μ ν eta_(mu nu)\eta_{\mu \nu}ημν in the neighborhood of P o P o P_(o)\mathscr{P}_{o}Po :
(8.15b) g μ ν , α ( P o ) = 0 (8.15b) g μ ν , α P o = 0 {:(8.15b)g_(mu nu,alpha)(P_(o))=0:}\begin{equation*} g_{\mu \nu, \alpha}\left(\mathscr{P}_{o}\right)=0 \tag{8.15b} \end{equation*}(8.15b)gμν,α(Po)=0
More tightly than this it cannot hold in general [ g μ ν , α β ( P o ) g μ ν , α β P o [g_(mu nu,alpha beta)(P_(o)):}\left[g_{\mu \nu, \alpha \beta}\left(\mathscr{P}_{o}\right)\right.[gμν,αβ(Po) cannot be set to zero]; spacetime curvature forces it to change. [Combine § 11.5 § 11.5 §11.5\S 11.5§11.5 with equations (8.24) and (8.44).]
Equivalence principle as vehicle between flat spacetime and curved
Local Lorentz frame: mathematical representation
Parallel transport defined
Covariant derivative defined
Gradient defined
Connection coefficients defined
An observer in a local Lorentz frame in curved spacetime can compare vectors and tensors at neighboring events, just as he would in flat spacetime. But to make the comparison, he must parallel-transport them to a common event. For him the act of parallel transport is simple: he keeps all Lorentz-frame components fixed, just as if he were in flat spacetime. But for a man without a local Lorentz frameperhaps with no coordinate system or basis vectors at all-parallel transport is less trivial. He must either ask his Lorentz-based friend the result, or he must use a more sophisticated technique. One technique he can use-a "Schild's ladder" construction that requires no coordinates or basis vectors whatsoever-is described in § 10.2 § 10.2 §10.2\S 10.2§10.2 and Box 10.2. But the Track-1 reader need not master Schild's ladder. He can always ask a local Lorentz observer what the result of any given parallel transport is, or he can use general formulas worked out below.
Comparison by parallel transport is the foundation on which rests the gradient of a tensor field, T T grad T\boldsymbol{\nabla} \boldsymbol{T}T. No mention of parallel transport was made in § 3.5 § 3.5 §3.5\S 3.5§3.5, where the gradient was first defined, but parallel transport occurred implicitly: one defined T T grad T\boldsymbol{\nabla} \boldsymbol{T}T in such a way that its components were T α β , γ = T α β / x γ [ T α β , γ = T α β / x γ T^(alpha)_(beta,gamma)=delT^(alpha)_(beta)//delx^(gamma)[:}T^{\alpha}{ }_{\beta, \gamma}=\partial T^{\alpha}{ }_{\beta} / \partial x^{\gamma}\left[\right.Tαβ,γ=Tαβ/xγ[ for T T T\boldsymbol{T}T a ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11) tensor] ] ] ]]]; i.e., one asked T T grad T\boldsymbol{\nabla} \boldsymbol{T}T to measure how much the Lorentz-frame components of T T T\boldsymbol{T}T change from point to point. But "no change in Lorentz components" would have meant "parallel transport," so one was implicitly asking for the change in T T T\boldsymbol{T}T relative to what T T T\boldsymbol{T}T would have been after pure parallel transport.
To codify in abstract notation this concept of differentiation, proceed as follows. First define the "covariant derivative" u T u T grad_(u)T\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}uT of T T T\boldsymbol{T}T along a curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ), whose tangent vector is u = d P / d λ u = d P / d λ u=dP//d lambda\boldsymbol{u}=d \mathscr{P} / d \lambdau=dP/dλ :
(8.16) ( u T ) at P ( 0 ) = Lim ε 0 { T [ P ( ε ) ] parallel-transported to P ( 0 ) T [ P ( 0 ) ] ε } . (8.16) u T at P ( 0 ) = Lim ε 0 T [ P ( ε ) ] parallel-transported to  P ( 0 ) T [ P ( 0 ) ] ε . {:(8.16)(grad_(u)T)_(atP(0))=Lim_(epsi rarr0){(T[P(epsi)]_("parallel-transported to "P(0))-T[P(0)])/(epsi)}.:}\begin{equation*} \left(\boldsymbol{\nabla}_{u} \boldsymbol{T}\right)_{\mathrm{at} \mathscr{P}(0)}=\operatorname{Lim}_{\varepsilon \rightarrow 0}\left\{\frac{\boldsymbol{T}[\mathscr{P}(\varepsilon)]_{\text {parallel-transported to } \mathscr{P}(0)}-\boldsymbol{T}[\mathscr{P}(0)]}{\varepsilon}\right\} . \tag{8.16} \end{equation*}(8.16)(uT)atP(0)=Limε0{T[P(ε)]parallel-transported to P(0)T[P(0)]ε}.
(See Figure 8.2 for the special case where T T T\boldsymbol{T}T is a vector field v v v\boldsymbol{v}v.) Then define T T grad T\boldsymbol{\nabla} \boldsymbol{T}T to be the linear machine, that gives u T u T grad_(u)T\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}uT when u u u\boldsymbol{u}u is inserted into its last slot:
(8.17) T ( , , u ) u T . (8.17) T ( , , u ) u T . {:(8.17)grad T(dots","dots","u)-=grad_(u)T.:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{T}(\ldots, \ldots, u) \equiv \boldsymbol{\nabla}_{u} \boldsymbol{T} . \tag{8.17} \end{equation*}(8.17)T(,,u)uT.
The result is the same animal ("gradient") as was defined in § 3.5 § 3.5 §3.5\S 3.5§3.5 (for proof see exercise 8.8). But this alternative definition makes clear the relationship to parallel transport, including the fact that
(8.18) u T = 0 T is parallel-transported along u = d P / d λ . (8.18) u T = 0 T  is parallel-transported along  u = d P / d λ . {:(8.18)grad_(u)T=0Longleftrightarrow T" is parallel-transported along "u=dP//d lambda.:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}=0 \Longleftrightarrow \boldsymbol{T} \text { is parallel-transported along } \boldsymbol{u}=d \mathscr{P} / d \lambda . \tag{8.18} \end{equation*}(8.18)uT=0T is parallel-transported along u=dP/dλ.
In a local Lorentz frame, the components of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T are directional derivatives of the components of T : T β α , γ T : T β α , γ T:T^(beta)_(alpha,gamma)\boldsymbol{T}: T^{\beta}{ }_{\alpha, \gamma}T:Tβα,γ. Not so in a general basis. If { e β ( P ) } e β ( P ) {e_(beta)(P)}\left\{\boldsymbol{e}_{\beta}(\mathscr{P})\right\}{eβ(P)} is a basis that varies arbitrarily but smoothly from point to point, and { ω α ( P ) } ω α ( P ) {omega^(alpha)(P)}\left\{\boldsymbol{\omega}^{\alpha}(\mathscr{P})\right\}{ωα(P)} is its dual basis, then T = ( T β α e β ω α ) T = T β α e β ω α grad T=grad(T^(beta)_(alpha)e_(beta)oxomega^(alpha))\boldsymbol{\nabla} \boldsymbol{T}=\boldsymbol{\nabla}\left(T^{\beta}{ }_{\alpha} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}\right)T=(Tβαeβωα) will contain contributions from e β e β grade_(beta)\boldsymbol{\nabla} \boldsymbol{e}_{\beta}eβ and ω α ω α gradomega^(alpha)\boldsymbol{\nabla} \boldsymbol{\omega}^{\alpha}ωα, as well as from T β α d T β α = T β α , γ ω γ T β α d T β α = T β α , γ ω γ gradT^(beta)_(alpha)-=dT^(beta)_(alpha)=T^(beta)_(alpha,gamma)omega^(gamma)\boldsymbol{\nabla} T^{\beta}{ }_{\alpha} \equiv \boldsymbol{d} T^{\beta}{ }_{\alpha}=T^{\beta}{ }_{\alpha, \gamma} \boldsymbol{\omega}^{\gamma}TβαdTβα=Tβα,γωγ.
To quantify the contributions from e β e β grade_(beta)\boldsymbol{\nabla} \boldsymbol{e}_{\beta}eβ and ω α ω α gradomega^(alpha)\boldsymbol{\nabla} \boldsymbol{\omega}^{\alpha}ωα, i.e., to quantify the twisting, turning, expansion, and contraction of the basis vectors and 1-forms, one defines "connection coefficients":
Figure 8.2.
Definition of the covariant derivative " u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv " of a vector field v v v\boldsymbol{v}v along a curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ), with tangent vector u d F / d λ u d F / d λ u-=dF//d lambda\boldsymbol{u} \equiv d \mathscr{F} / d \lambdaudF/dλ : (1) choose a point P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0) on the curve, at which to evaluate u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv. (2) Choose a nearby point P ( ε ) P ( ε ) P(epsi)\mathscr{P}(\varepsilon)P(ε) on the curve. (3) Parallel-transport v [ P ( ε ) ] v [ P ( ε ) ] v[P(epsi)]\boldsymbol{v}[\mathscr{P}(\varepsilon)]v[P(ε)] along the curve back to P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0), getting the vector v i [ P ( ε ) ] v i [ P ( ε ) ] v_(i)[P(epsi)]\boldsymbol{v}_{i}[\mathscr{P}(\varepsilon)]vi[P(ε)]. (4) Take the difference δ v v il [ P ( ε ) ] v [ P ( 0 ) ] δ v v il [ P ( ε ) ] v [ P ( 0 ) ] delta v-=v_(il)[P(epsi)]-v[P(0)]\delta \boldsymbol{v} \equiv \boldsymbol{v}_{\mathrm{il}}[\mathscr{P}(\varepsilon)]-\boldsymbol{v}[\mathscr{P}(0)]δvvil[P(ε)]v[P(0)]. (5) Then u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv is defined by
u v Lim t 0 δ v ε = Lim ε 0 { v t [ P ( ε ) ] v [ P ( 0 ) ] ε } u v Lim t 0 δ v ε = Lim ε 0 v t [ P ( ε ) ] v [ P ( 0 ) ] ε grad_(u)v-=Lim_(t rarr0)(delta v)/(epsi)=Lim_(epsi rarr0){(v_(t)[P(epsi)]-v[P(0)])/(epsi)}\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v} \equiv \operatorname{Lim}_{t \rightarrow 0} \frac{\delta \boldsymbol{v}}{\varepsilon}=\operatorname{Lim}_{\varepsilon \rightarrow 0}\left\{\frac{\boldsymbol{v}_{\mathrm{t}}[\mathscr{P}(\varepsilon)]-\boldsymbol{v}[\mathscr{P}(0)]}{\varepsilon}\right\}uvLimt0δvε=Limε0{vt[P(ε)]v[P(0)]ε}
(8.19a) Γ β γ α ω α , γ e β [ Note reversal of β and γ to make the differentiating index come last on Γ ] ↑≡ e γ ] (8.19a) Γ β γ α ω α , γ e β  Note reversal of  β  and  γ  to make the   differentiating index come last on  Γ ↑≡ e γ {:[(8.19a)Gamma_(beta gamma)^(alpha)-=(:omega^(alpha),grad_(gamma)e_(beta):)quad[[" Note reversal of "beta" and "gamma" to make the "],[" differentiating index come last on "Gamma]]],[{: uarr -=grad_(e_(gamma))]]:}\begin{align*} & \Gamma_{\beta \gamma}^{\alpha} \equiv\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{\nabla}_{\gamma} \boldsymbol{e}_{\beta}\right\rangle \quad\left[\begin{array}{c} \text { Note reversal of } \beta \text { and } \gamma \text { to make the } \\ \text { differentiating index come last on } \Gamma \end{array}\right] \tag{8.19a}\\ &\left.\uparrow \equiv \boldsymbol{\nabla}_{\boldsymbol{e}_{\gamma}}\right] \end{align*}(8.19a)Γβγαωα,γeβ[ Note reversal of β and γ to make the  differentiating index come last on Γ]↑≡eγ]
and one proves (exercise 8.12) that
(8.19b) γ ω α , e β = Γ β γ α . (8.19b) γ ω α , e β = Γ β γ α . {:(8.19b)(:grad_(gamma)omega^(alpha),e_(beta):)=-Gamma_(beta gamma)^(alpha).:}\begin{equation*} \left\langle\boldsymbol{\nabla}_{\gamma} \boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=-\Gamma_{\beta \gamma}^{\alpha} . \tag{8.19b} \end{equation*}(8.19b)γωα,eβ=Γβγα.
In terms of these coefficients and
(8.20) T α , γ β γ T α β e γ T α β γ T α β (8.20) T α , γ β γ T α β e γ T α β γ T α β {:(8.20)T_(alpha,gamma)^(beta)-=grad_(gamma)T_(alpha)^(beta)-=del_(e_(gamma))T_(alpha)^(beta)-=del_(gamma)T_(alpha)^(beta):}\begin{equation*} T_{\alpha, \gamma}^{\beta} \equiv \nabla_{\gamma} T_{\alpha}^{\beta} \equiv \partial_{e_{\gamma}} T_{\alpha}^{\beta} \equiv \partial_{\gamma} T_{\alpha}^{\beta} \tag{8.20} \end{equation*}(8.20)Tα,γβγTαβeγTαβγTαβ
the components of the gradient, denoted T β α ; γ T β α ; γ T^(beta)_(alpha;gamma)T^{\beta}{ }_{\alpha ; \gamma}Tβα;γ, are
(8.21) T α ; γ β = T α , γ β + Γ μ γ β T α μ Γ α γ μ T μ β (8.21) T α ; γ β = T α , γ β + Γ μ γ β T α μ Γ α γ μ T μ β {:(8.21)T_(alpha;gamma)^(beta)=T_(alpha,gamma)^(beta)+Gamma_(mu gamma)^(beta)T_(alpha)^(mu)-Gamma_(alpha gamma)^(mu)T_(mu)^(beta):}\begin{equation*} T_{\alpha ; \gamma}^{\beta}=T_{\alpha, \gamma}^{\beta}+\Gamma_{\mu \gamma}^{\beta} T_{\alpha}^{\mu}-\Gamma_{\alpha \gamma}^{\mu} T_{\mu}^{\beta} \tag{8.21} \end{equation*}(8.21)Tα;γβ=Tα,γβ+ΓμγβTαμΓαγμTμβ
(see exercise 8.13). If the basis at the event where T T grad T\boldsymbol{\nabla} \boldsymbol{T}T is calculated were a local Lorentz frame, the components of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T would just be T β α , γ T β α , γ T^(beta)_(alpha,gamma)T^{\beta}{ }_{\alpha, \gamma}Tβα,γ. Because it is not, one must correct this "Lorentz-frame" value for the twisting, turning, expansion, and contraction of the basis vectors and 1 -forms. The " Γ T Γ T Gamma T\Gamma TΓT " terms in equation (8.21) are the necessary corrections-one for each index of T T T\boldsymbol{T}T. The pattern of these correction terms is easy to remember: (1) " + " sign if index being corrected is up, " - " sign if it is down; (2) differentiation index ( γ γ gamma\gammaγ in above case) always at end of Γ Γ Gamma\GammaΓ; (3) index being corrected ( β β beta\betaβ in first term, α α alpha\alphaα in second) shifts from T T TTT onto Γ Γ Gamma\GammaΓ and gets replaced on T T TTT by a dummy summation index ( μ ) ( μ ) (mu)(\mu)(μ).
  • = = === - = = === -
Components of gradient in arbitrary frame
Components of covariant derivative
Calculation of connection coefficients from metric and commutators
Knowing the components (8.21) of the gradient, one can calculate the components of the covariant derivative u T u T grad_(u)T\boldsymbol{\nabla}_{u} \boldsymbol{T}uT by a simple contraction into u γ u γ u^(gamma)u^{\gamma}uγ [see equation (8.17)]:
(8.22) u T = ( T β α ; γ u γ ) e β ω α . (8.22) u T = T β α ; γ u γ e β ω α . {:(8.22)grad_(u)T=(T^(beta)_(alpha;gamma)u^(gamma))e_(beta)oxomega^(alpha).:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}=\left(T^{\beta}{ }_{\alpha ; \gamma} u^{\gamma}\right) \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha} . \tag{8.22} \end{equation*}(8.22)uT=(Tβα;γuγ)eβωα.
When u u u\boldsymbol{u}u is the tangent vector to a curve P ( λ ) , u = d P / d λ P ( λ ) , u = d P / d λ P(lambda),u=dP//d lambda\mathscr{P}(\lambda), \boldsymbol{u}=d \mathscr{P} / d \lambdaP(λ),u=dP/dλ, one uses the notation D T β α / d λ D T β α / d λ DT^(beta)_(alpha)//d lambdaD T^{\beta}{ }_{\alpha} / d \lambdaDTβα/dλ for the components of u T u T grad_(u)T\boldsymbol{\nabla}_{u} \boldsymbol{T}uT :
D T β α d λ T β α ; γ u γ = ψ T β α ; γ d x γ d λ (8.23) = ( T β α , γ + " Γ T corrections ) d x γ / d λ = d T β α d λ + ( Γ β μ γ T μ α Γ μ α γ T β μ ) d x γ d λ . D T β α d λ T β α ; γ u γ = ψ T β α ; γ d x γ d λ (8.23) = T β α , γ + " Γ T  corrections  d x γ / d λ = d T β α d λ + Γ β μ γ T μ α Γ μ α γ T β μ d x γ d λ . {:[(DT^(beta)_(alpha))/(d lambda)-=T^(beta)_(alpha;gamma)u^(gamma)=^(psi)T^(beta)_(alpha;gamma)(dx^(gamma))/(d lambda)],[(8.23)=(T^(beta)_(alpha,gamma)+"GammaT^('')" corrections ")dx^(gamma)//d lambda],[=(dT^(beta)_(alpha))/(d lambda)+(Gamma^(beta)_(mu gamma)T^(mu)_(alpha)-Gamma^(mu)_(alpha gamma)T^(beta)_(mu))(dx^(gamma))/(d lambda).]:}\begin{align*} \frac{D T^{\beta}{ }_{\alpha}}{d \lambda} & \equiv T^{\beta}{ }_{\alpha ; \gamma} u^{\gamma} \stackrel{\psi}{=} T^{\beta}{ }_{\alpha ; \gamma} \frac{d x^{\gamma}}{d \lambda} \\ & =\left(T^{\beta}{ }_{\alpha, \gamma}+" \Gamma T^{\prime \prime} \text { corrections }\right) d x^{\gamma} / d \lambda \tag{8.23}\\ & =\frac{d T^{\beta}{ }_{\alpha}}{d \lambda}+\left(\Gamma^{\beta}{ }_{\mu \gamma} T^{\mu}{ }_{\alpha}-\Gamma^{\mu}{ }_{\alpha \gamma} T^{\beta}{ }_{\mu}\right) \frac{d x^{\gamma}}{d \lambda} . \end{align*}DTβαdλTβα;γuγ=ψTβα;γdxγdλ(8.23)=(Tβα,γ+"ΓT corrections )dxγ/dλ=dTβαdλ+(ΓβμγTμαΓμαγTβμ)dxγdλ.
The ";" in T β α ; γ T β α ; γ T^(beta)_(alpha;gamma)T^{\beta}{ }_{\alpha ; \gamma}Tβα;γ reminds one to correct T β α , γ T β α , γ T^(beta)_(alpha,gamma)T^{\beta}{ }_{\alpha, \gamma}Tβα,γ with " Γ T Γ T GammaT^('')\Gamma T^{\prime \prime}ΓT terms; similarly, the " D D DDD " in D T β α / d λ D T β α / d λ DT^(beta)_(alpha)//d lambdaD T^{\beta}{ }_{\alpha} / d \lambdaDTβα/dλ reminds one to correct d T β α / d λ d T β α / d λ dT^(beta)_(alpha)//d lambdad T^{\beta}{ }_{\alpha} / d \lambdadTβα/dλ with " Γ T Γ T Gamma T\Gamma TΓT " terms.
This is all well and good, but how does one find out the connection coefficients Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ for a given basis? The answer is derived in exercise 8.15. It says: (1) take the metric coefficients in the given basis; (2) calculate their directional derivatives along the basis directions
(8.24a) g β γ , μ μ g β γ = g β γ / x μ ; [ e μ ] [if a coordinate basis, e μ = P / x μ , is being used] (8.24a) g β γ , μ μ g β γ = g β γ / x μ ; e μ  [if a coordinate basis,  e μ = P / x μ ,  is being used]  {:[(8.24a)g_(beta gamma,mu)-=-=del_(mu)g_(beta gamma)=delg_(beta gamma)//delx^(mu);],[uarr_([-=del_(e_(mu))])uarr" [if a coordinate basis, "e_(mu)=delP//delx^(mu)","" is being used] "]:}\begin{align*} g_{\beta \gamma, \mu} \equiv & \equiv \partial_{\mu} g_{\beta \gamma}=\partial g_{\beta \gamma} / \partial x^{\mu} ; \tag{8.24a}\\ & \uparrow_{\left[\equiv \partial_{\boldsymbol{e}_{\mu}}\right]} \uparrow \text { [if a coordinate basis, } \boldsymbol{e}_{\mu}=\partial \mathscr{P} / \partial x^{\mu}, \text { is being used] } \end{align*}(8.24a)gβγ,μμgβγ=gβγ/xμ;[eμ] [if a coordinate basis, eμ=P/xμ, is being used] 
(3) calculate the commutation coefficients of the basis [equations (8.14) in general; c μ β γ = 0 c μ β γ = 0 c_(mu beta gamma)=0c_{\mu \beta \gamma}=0cμβγ=0 in special case of coordinate basis]; (4) calculate the "covariant connection coefficients"
(8.24b) Γ μ β γ = 1 2 ( g μ β , γ + g μ γ , β g β γ , μ + c μ β γ + c μ γ β c β γ μ β γ ) ] (8.24b) Γ μ β γ = 1 2 ( g μ β , γ + g μ γ , β g β γ , μ + c μ β γ + c μ γ β c β γ μ β γ ) {:(8.24b){:Gamma_(mu beta gamma)=(1)/(2)(g_(mu beta,gamma)+g_(mu gamma,beta)-g_(beta gamma,mu)+ubrace(c_(mu beta gamma)+c_(mu gamma beta)-c_(beta gamma mu)ubrace)_(uarr beta gamma))]:}\left.\begin{array}{c} \Gamma_{\mu \beta \gamma}=\frac{1}{2}(g_{\mu \beta, \gamma}+g_{\mu \gamma, \beta}-g_{\beta \gamma, \mu}+\underbrace{c_{\mu \beta \gamma}+c_{\mu \gamma \beta}-c_{\beta \gamma \mu}}_{\uparrow \beta \gamma}) \tag{8.24b} \end{array}\right](8.24b)Γμβγ=12(gμβ,γ+gμγ,βgβγ,μ+cμβγ+cμγβcβγμβγ)]
(5) raise an index to get the connection coefficients:
(8.24c) Γ β γ α = g α μ Γ μ β γ (8.24c) Γ β γ α = g α μ Γ μ β γ {:(8.24c)Gamma_(beta gamma)^(alpha)=g^(alpha mu)Gamma_(mu beta gamma):}\begin{equation*} \Gamma_{\beta \gamma}^{\alpha}=g^{\alpha \mu} \Gamma_{\mu \beta \gamma} \tag{8.24c} \end{equation*}(8.24c)Γβγα=gαμΓμβγ
[Note on terminology: a coordinate basis always has c α β γ = 0 c α β γ = 0 c_(alpha beta gamma)=0c_{\alpha \beta \gamma}=0cαβγ=0, and is sometimes called holonomic; a noncoordinate basis always has some of its c α β γ c α β γ c_(alpha beta gamma)c_{\alpha \beta \gamma}cαβγ nonzero, and is sometimes called anholonomic. In the holonomic case, the connection coefficients are sometimes called Christoffel symbols.]
The component notation, with its semicolons, commas, D D DDD 's, connection coefficients, etc., looks rather formidable at first. But it bears great computational power, one discovers as one proceeds deep into gravitation theory; and its rules of manipulation
are simple enough to be learned easily. By contrast, the abstract notation ( T , u T T , u T (grad T,grad_(u)T:}\left(\boldsymbol{\nabla} \boldsymbol{T}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}\right.(T,uT, etc.) is poorly suited to complex calculations; but it possesses great conceptual power.
This contrast shows clearly in the way the two notations handle the concept of geodesic. A geodesic of spacetime is a curve that is straight and uniformly parametrized, as measured in each local Lorentz frame along its way. If the geodesic is timelike, then it is a possible world line for a freely falling particle, and its uniformly ticking parameter λ λ lambda\lambdaλ (called "affine parameter") is a multiple of the particle's proper time, λ = a τ + b λ = a τ + b lambda=a tau+b\lambda=a \tau+bλ=aτ+b. (Principle of equivalence: test particles move on straight lines in local Lorentz frames, and each particle's clock ticks at a uniform rate as measured by any Lorentz observer.) This definition of geodesic is readily translated into abstract, coordinate-free language: a geodesic is a curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) that parallel-transports its tangent vector u = d P / d λ u = d P / d λ u=dP//d lambda\boldsymbol{u}=d \mathscr{P} / d \lambdau=dP/dλ along itself-
(8.25) u u = 0 (8.25) u u = 0 {:(8.25)grad_(u)u=0:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0 \tag{8.25} \end{equation*}(8.25)uu=0
(See Figure 10.1.) What could be simpler conceptually? But to compute the geodesic, given an initial event P o P o P_(o)\mathscr{P}_{o}Po and initial tangent vector u ( O o ) u O o u(O_(o))\boldsymbol{u}\left(\mathscr{O}_{o}\right)u(Oo) there, one must use the component formalism. Introduce a coordinate system x α ( P ) x α ( P ) x^(alpha)(P)x^{\alpha}(\mathscr{P})xα(P), in which u α = d x α / d λ u α = d x α / d λ u^(alpha)=dx^(alpha)//d lambdau^{\alpha}=d x^{\alpha} / d \lambdauα=dxα/dλ, and write the component version of equation (8.25) as
0 = D ( d x α / d λ ) d λ = d ( d x α / d λ ) d λ + ( Γ α μ γ d x μ d λ ) d x γ d λ 0 = D d x α / d λ d λ = d d x α / d λ d λ + Γ α μ γ d x μ d λ d x γ d λ 0=(D(dx^(alpha)//d lambda))/(d lambda)=(d(dx^(alpha)//d lambda))/(d lambda)+(Gamma^(alpha)_(mu gamma)(dx^(mu))/(d lambda))(dx^(gamma))/(d lambda)0=\frac{D\left(d x^{\alpha} / d \lambda\right)}{d \lambda}=\frac{d\left(d x^{\alpha} / d \lambda\right)}{d \lambda}+\left(\Gamma^{\alpha}{ }_{\mu \gamma} \frac{d x^{\mu}}{d \lambda}\right) \frac{d x^{\gamma}}{d \lambda}0=D(dxα/dλ)dλ=d(dxα/dλ)dλ+(Γαμγdxμdλ)dxγdλ
[see equation (8.23), with one less index on T T TTT ]; i.e.,
(8.26) d 2 x α d λ 2 + Γ α μ γ d x μ d λ d x γ d λ = 0 (8.26) d 2 x α d λ 2 + Γ α μ γ d x μ d λ d x γ d λ = 0 {:(8.26)(d^(2)x^(alpha))/(dlambda^(2))+Gamma^(alpha)_(mu gamma)(dx^(mu))/(d lambda)(dx^(gamma))/(d lambda)=0:}\begin{equation*} \frac{d^{2} x^{\alpha}}{d \lambda^{2}}+\Gamma^{\alpha}{ }_{\mu \gamma} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\gamma}}{d \lambda}=0 \tag{8.26} \end{equation*}(8.26)d2xαdλ2+Γαμγdxμdλdxγdλ=0
This geodesic equation can be solved (in principle) for the coordinates of the geodesic, x α ( λ ) x α ( λ ) x^(alpha)(lambda)x^{\alpha}(\lambda)xα(λ), when initial data [ x α x α [x^(alpha):}\left[x^{\alpha}\right.[xα and d x α / d λ d x α / d λ dx^(alpha)//d lambdad x^{\alpha} / d \lambdadxα/dλ at λ = λ 0 ] λ = λ 0 {: lambda=lambda_(0)]\left.\lambda=\lambda_{0}\right]λ=λ0] have been specified.
The geodesics of the Earth's surface (great circles) are a foil against which one can visualize connection coefficients; see Figure 8.3.
The material of this section is presented more deeply and from a different viewpoint in Chapters 10 and 13. The Track-2 reader who plans to study those chapters is advised to ignore the following exercises. The Track-1 reader who intends to skip Chapters 9 15 9 15 9-159-15915 will gain necessary experience with the component formalism by working exercises 8.4 8.7 8.4 8.7 8.4-8.78.4-8.78.48.7. Less important to him, but valuable nonetheless, are exercises 8.8 8.15 8.8 8.15 8.8-8.158.8-8.158.88.15, which develop the formalism of covariant derivatives and connection coefficients in a systematic manner. The most important results of these exercises will be summarized in Box 8.6 (pages 223 and 224).

Exercise 8.4. PRACTICE IN WRITING COMPONENTS OF GRADIENT

Rewrite the following quantities in terms of ordinary derivatives ( f , γ e f γ f f , γ e f γ f f_(,gamma)-=del_(e)f-=grad_(gamma)ff_{, \gamma} \equiv \partial_{\boldsymbol{e}} f \equiv \boldsymbol{\nabla}_{\gamma} ff,γefγf ) and " Γ T Γ T Gamma T\Gamma TΓT " correction terms: (a) T ; γ T ; γ T_(;gamma)T_{; \gamma}T;γ where T T TTT is a function. (b) T α ; γ T α ; γ T^(alpha)_(;gamma)T^{\alpha}{ }_{; \gamma}Tα;γ where T T T\boldsymbol{T}T is a vector. (c) T α ; γ T α ; γ T_(alpha;gamma)T_{\alpha ; \gamma}Tα;γ where T T T\boldsymbol{T}T is a 1 -form. (d) T α β δ ; γ T α β δ ; γ T^(alpha)_(betadelta^(**);gamma)T^{\alpha}{ }_{\beta \delta^{*} ; \gamma}Tαβδ;γ [Answer:
(a) T i γ = T , γ T i γ = T , γ T_(i gamma)=T_(,gamma)T_{i \gamma}=T_{, \gamma}Tiγ=T,γ.
(b) T α ; γ = T α , γ + Γ α μ γ T μ T α ; γ = T α , γ + Γ α μ γ T μ T^(alpha)_(;gamma)=T^(alpha)_(,gamma)+Gamma^(alpha)_(mu gamma)T^(mu)T^{\alpha}{ }_{; \gamma}=T^{\alpha}{ }_{, \gamma}+\Gamma^{\alpha}{ }_{\mu \gamma} T^{\mu}Tα;γ=Tα,γ+ΓαμγTμ.
(c) T α ; γ = T α , γ Γ μ α γ T μ T α ; γ = T α , γ Γ μ α γ T μ T_(alpha;gamma)=T_(alpha,gamma)-Gamma^(mu)_(alpha gamma)T_(mu)T_{\alpha ; \gamma}=T_{\alpha, \gamma}-\Gamma^{\mu}{ }_{\alpha \gamma} T_{\mu}Tα;γ=Tα,γΓμαγTμ.
(d) T α β δ e ; γ = T α β δ e , γ + Γ α μ γ T μ β δ e Γ μ β γ T α μ δ e Γ μ δ γ T α β μ + Γ e μ γ T α β δ μ . ] T α β δ e ; γ = T α β δ e , γ + Γ α μ γ T μ β δ e Γ μ β γ T α μ δ e Γ μ δ γ T α β μ + Γ e μ γ T α β δ μ . {:T^(alpha)_(beta delta)^(e);gamma=T^(alpha)_(beta delta)^(e),gamma+Gamma^(alpha)_(mu gamma)T^(mu)_(beta delta)^(e)-Gamma^(mu)_(beta gamma)T^(alpha)_(mu delta)^(e)-Gamma^(mu)_(delta gamma)T^(alpha)_(beta mu)^(ℓ)+Gamma^(e)_(mu gamma)T^(alpha)_(beta delta)^(mu).]\left.T^{\alpha}{ }_{\beta \delta}{ }^{e} ; \gamma=T^{\alpha}{ }_{\beta \delta}{ }^{e}, \gamma+\Gamma^{\alpha}{ }_{\mu \gamma} T^{\mu}{ }_{\beta \delta}{ }^{e}-\Gamma^{\mu}{ }_{\beta \gamma} T^{\alpha}{ }_{\mu \delta}{ }^{e}-\Gamma^{\mu}{ }_{\delta \gamma} T^{\alpha}{ }_{\beta \mu}{ }^{\ell}+\Gamma^{e}{ }_{\mu \gamma} T^{\alpha}{ }_{\beta \delta}{ }^{\mu}.\right]Tαβδe;γ=Tαβδe,γ+ΓαμγTμβδeΓμβγTαμδeΓμδγTαβμ+ΓeμγTαβδμ.]
Figure 8.3.
The why of connection coefficients, schematically portrayed. The aviator pursuing his great circle route from Peking to Vancouver finds himself early going north, but later going south, although he is navigating the straightest route that is at all open to him (geodesic). The apparent change in direction indicates a turning, not in his route, but in the system of coordinates with respect to which his route is described. The vector v v v\boldsymbol{v}v of his velocity (a vector defined not on spacetime but rather on the Earth's two-dimensional surface), carried forward by parallel transport from an earlier moment to a later moment, finds itself in agreement with the velocity that he is then pursuing; or, in the abstract language of coordinate-free differential geometry, the covariant derivative v v v v grad_(v)v\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{v}vv vanishes along the route ("equation of a geodesic"). Though v v v\boldsymbol{v}v is in this sense constant, the individual pieces of which the navigator considers this vector to be built, v = v θ e θ + v ϕ e ϕ v = v θ e θ + v ϕ e ϕ v=v^(theta)e_(theta)+v^(phi)e_(phi)\boldsymbol{v}=v^{\theta} \boldsymbol{e}_{\theta}+v^{\phi} \boldsymbol{e}_{\phi}v=vθeθ+vϕeϕ, are not constant.
In the language of components, the quantities v θ v θ v^(theta)v^{\theta}vθ and v ϕ v ϕ v^(phi)v^{\phi}vϕ are changing along the route at a rate that annuls the covariant derivative of v v v\mathbf{v}v; thus
v v = a = a ϕ e ϕ + a θ e θ = 0 v v = a = a ϕ e ϕ + a θ e θ = 0 grad_(v)v=a=a^(phi)e_(phi)+a^(theta)e_(theta)=0\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{v}=\boldsymbol{a}=a^{\phi} \boldsymbol{e}_{\phi}+a^{\theta} \boldsymbol{e}_{\theta}=0vv=a=aϕeϕ+aθeθ=0
or
0 = a θ = d v θ d t + Γ θ m n v m v n 0 = a ϕ = d v ϕ d t + Γ ϕ m n v m v n 0 = a θ = d v θ d t + Γ θ m n v m v n 0 = a ϕ = d v ϕ d t + Γ ϕ m n v m v n {:[0=a^(theta)=(dv^(theta))/(dt)+Gamma^(theta)_(mn)v^(m)v^(n)],[0=a^(phi)=(dv^(phi))/(dt)+Gamma^(phi)_(mn)v^(m)v^(n)]:}\begin{aligned} & 0=a^{\theta}=\frac{d v^{\theta}}{d t}+\Gamma^{\theta}{ }_{m n} v^{m} v^{n} \\ & 0=a^{\phi}=\frac{d v^{\phi}}{d t}+\Gamma^{\phi}{ }_{m n} v^{m} v^{n} \end{aligned}0=aθ=dvθdt+Γθmnvmvn0=aϕ=dvϕdt+Γϕmnvmvn
In this sense the connection coefficients Γ j m n Γ j m n Gamma^(j)_(mn)\Gamma^{j}{ }_{m n}Γjmn serve as "turning coefficients" to tell how fast to "turn" the components of a vector in order to keep that vector constant (against the turning influence of the base vectors).
Alternatively, the navigator can use an "automatic pilot system" which parallel-transports its own base vectors along the plane's route:
v e θ = v e o = 0 v e θ = v e o = 0 grad_(v)e_(theta^('))=grad_(v)e_(o^('))=0\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{e}_{\theta^{\prime}}=\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{e}_{\mathbf{o}^{\prime}}=0veθ=veo=0
solid vectors at a a aaa become dotted vectors at B B B\mathscr{B}B. Then the components of v v v\boldsymbol{v}v must be kept fixed to achieve a great-circle route,
d v θ d t = d v ϕ d t = 0 ; d v θ d t = d v ϕ d t = 0 ; (dv^(theta^(')))/(dt)=(dv^(phi^(')))/(dt)=0;\frac{d v^{\theta^{\prime}}}{d t}=\frac{d v^{\phi^{\prime}}}{d t}=0 ;dvθdt=dvϕdt=0;
and the turning coefficients are used to describe the turning of the lines of latitude and longitude relative to this parallel-transported basis:
v e θ = e m I m θ n v n , v e ϕ = e m I m ϕ n v n . v e θ = e m I m θ n v n , v e ϕ = e m I m ϕ n v n . {:[grad_(v)e_(theta)=e_(m)I^(m)_(theta n)v^(n)","],[grad_(v)e_(phi)=e_(m)I^(m)_(phi n)v^(n).]:}\begin{aligned} \boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{e}_{\theta} & =\boldsymbol{e}_{m} I^{m}{ }_{\theta n} v^{n}, \\ \boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{e}_{\phi} & =\boldsymbol{e}_{m} I^{m}{ }_{\phi n} v^{n} . \end{aligned}veθ=emImθnvn,veϕ=emImϕnvn.
The same turning coefficients enter into both viewpoints. The only difference is in how these coefficients are used.

Exercise 8.5. A SHEET OF PAPER IN POLAR COORDINATES

The two-dimensional metric for a flat sheet of paper in polar coordinates ( r , θ ) ( r , θ ) (r,theta)(r, \theta)(r,θ) is d s 2 = d r 2 d s 2 = d r 2 ds^(2)=dr^(2)d s^{2}=d r^{2}ds2=dr2 + r 2 d ϕ 2 + r 2 d ϕ 2 +r^(2)dphi^(2)+r^{2} d \phi^{2}+r2dϕ2-or, in modern notation, g = d r d r + r 2 d ϕ d ϕ g = d r d r + r 2 d ϕ d ϕ g=dr ox dr+r^(2)d phi ox d phi\boldsymbol{g}=\boldsymbol{d} r \otimes \boldsymbol{d} r+r^{2} \boldsymbol{d} \phi \otimes \boldsymbol{d} \phig=drdr+r2dϕdϕ.
(a) Calculate the connection coefficients using equations (8.24). [Answer: Γ r ϕ ϕ = r Γ r ϕ ϕ = r Gamma^(r)_(phi phi)=-r\Gamma^{r}{ }_{\phi \phi}=-rΓrϕϕ=r; Γ ϕ r ϕ = Γ ϕ ϕ r = 1 / r Γ ϕ r ϕ = Γ ϕ ϕ r = 1 / r Gamma^(phi)_(r phi)=Gamma^(phi)_(phi r)=1//r\Gamma^{\phi}{ }_{r \phi}=\Gamma^{\phi}{ }_{\phi r}=1 / rΓϕrϕ=Γϕϕr=1/r; all others vanish.]
(b) Write down the geodesic equation in ( r , ϕ ) ( r , ϕ ) (r,phi)(r, \phi)(r,ϕ) coordinates. [Answer: d 2 r / d λ 2 d 2 r / d λ 2 d^(2)r//dlambda^(2)-d^{2} r / d \lambda^{2}-d2r/dλ2 r ( d ϕ ˙ / d λ ) 2 = 0 ; d 2 ϕ / d λ 2 + ( 2 / r ) ( d r / d λ ) ( d ϕ / d λ ) = 0 r ( d ϕ ˙ / d λ ) 2 = 0 ; d 2 ϕ / d λ 2 + ( 2 / r ) ( d r / d λ ) ( d ϕ / d λ ) = 0 r(dphi^(˙)//d lambda)^(2)=0;d^(2)phi//dlambda^(2)+(2//r)(dr//d lambda)(d phi//d lambda)=0r(d \dot{\phi} / d \lambda)^{2}=0 ; d^{2} \phi / d \lambda^{2}+(2 / r)(d r / d \lambda)(d \phi / d \lambda)=0r(dϕ˙/dλ)2=0;d2ϕ/dλ2+(2/r)(dr/dλ)(dϕ/dλ)=0.]
(c) Solve this geodesic equation for r ( λ ) r ( λ ) r(lambda)r(\lambda)r(λ) and ϕ ( λ ) ϕ ( λ ) phi(lambda)\phi(\lambda)ϕ(λ), and show that the solution is a uniformly parametrized straight line ( x r cos ϕ = a λ + b ( x r cos ϕ = a λ + b (x-=r cos phi=a lambda+b(x \equiv r \cos \phi=a \lambda+b(xrcosϕ=aλ+b for some a a aaa and b ; y r sin ϕ = j λ + k b ; y r sin ϕ = j λ + k b;y-=r sin phi=j lambda+kb ; y \equiv r \sin \phi=j \lambda+kb;yrsinϕ=jλ+k for some j j jjj and k k kkk ).
(d) Verify that the noncoordinate basis e r ^ e r = P / r , e ϕ ^ r 1 e ϕ r 1 P / ϕ e r ^ e r = P / r , e ϕ ^ r 1 e ϕ r 1 P / ϕ e_( hat(r))-=e_(r)=delP//del r,e_( hat(phi))-=r^(-1)e_(phi)quadr^(-1)delP//del phi\boldsymbol{e}_{\hat{r}} \equiv \boldsymbol{e}_{r}=\partial \mathscr{P} / \partial r, \boldsymbol{e}_{\hat{\phi}} \equiv r^{-1} \boldsymbol{e}_{\phi} \quad r^{-1} \partial \mathscr{P} / \partial \phier^er=P/r,eϕ^r1eϕr1P/ϕ, ω r = d r , ω ϕ ^ = r d ϕ ω r = d r , ω ϕ ^ = r d ϕ omega^(r)=dr,omega^( hat(phi))=rd phi\boldsymbol{\omega}^{r}=\boldsymbol{d} r, \boldsymbol{\omega}^{\hat{\phi}}=r \boldsymbol{d} \phiωr=dr,ωϕ^=rdϕ is orthonormal, and that ω α , e β ^ = δ α ^ ω α , e β ^ = δ α ^ (:omega^(alpha),e_( hat(beta)):)=delta^( hat(alpha))\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\hat{\beta}}\right\rangle=\delta^{\hat{\alpha}}ωα,eβ^=δα^. . Then calculate the connection coefficients of this basis from a knowledge [part (a)] of the connection of the coordinate basis. [Answer:
Γ ϕ ^ ϕ ^ r ^ = ω ϕ ^ , r ^ e ϕ ^ = r d ϕ ^ , r ( r 1 e ϕ ) = r d ϕ , ( r r 1 ) e ϕ + r 1 ( r e ϕ ) = r d ϕ , r 2 e ϕ + d ϕ , r e ϕ = r 1 + Γ ϕ ϕ r = r 1 + r 1 = 0 ; Γ ϕ ^ ϕ ^ r ^ = ω ϕ ^ , r ^ e ϕ ^ = r d ϕ ^ , r r 1 e ϕ = r d ϕ , r r 1 e ϕ + r 1 r e ϕ = r d ϕ , r 2 e ϕ + d ϕ , r e ϕ = r 1 + Γ ϕ ϕ r = r 1 + r 1 = 0 ; {:[Gamma^( hat(phi)_( hat(phi) hat(r)))=(:omega^( hat(phi)),grad_( hat(r))e_( hat(phi)):)=(:rd( hat(phi)),grad_(r)(r^(-1)e_(phi)):}],[=r(:d phi,(grad_(r)r^(-1))e_(phi)+r^(-1)(grad_(r)e_(phi)):)=r(:d phi,-r^(-2)e_(phi):)+(:d phi,grad_(r)e_(phi):)],[=-r^(-1)+Gamma^(phi)_(phi r)=-r^(-1)+r^(-1)=0;]:}\begin{aligned} \Gamma^{\hat{\phi}_{\hat{\phi} \hat{r}}} & =\left\langle\boldsymbol{\omega}^{\hat{\phi}}, \boldsymbol{\nabla}_{\hat{r}} \boldsymbol{e}_{\hat{\phi}}\right\rangle=\left\langle r \boldsymbol{d} \hat{\phi}, \boldsymbol{\nabla}_{r}\left(r^{-1} \boldsymbol{e}_{\phi}\right)\right. \\ & =r\left\langle\boldsymbol{d} \phi,\left(\boldsymbol{\nabla}_{r} r^{-1}\right) \boldsymbol{e}_{\phi}+r^{-1}\left(\boldsymbol{\nabla}_{r} \boldsymbol{e}_{\phi}\right)\right\rangle=r\left\langle\boldsymbol{d} \phi,-r^{-2} \boldsymbol{e}_{\phi}\right\rangle+\left\langle\boldsymbol{d} \phi, \boldsymbol{\nabla}_{r} \boldsymbol{e}_{\phi}\right\rangle \\ & =-r^{-1}+\Gamma^{\phi}{ }_{\phi r}=-r^{-1}+r^{-1}=0 ; \end{aligned}Γϕ^ϕ^r^=ωϕ^,r^eϕ^=rdϕ^,r(r1eϕ)=rdϕ,(rr1)eϕ+r1(reϕ)=rdϕ,r2eϕ+dϕ,reϕ=r1+Γϕϕr=r1+r1=0;
similarly, Γ ϕ ^ r ^ ϕ ^ = + 1 / r , Γ r ^ ϕ ^ ϕ ^ = 1 / r Γ ϕ ^ r ^ ϕ ^ = + 1 / r , Γ r ^ ϕ ^ ϕ ^ = 1 / r Gamma^( hat(phi)_( hat(r) hat(phi)))=+1//r,Gamma^( hat(r)_( hat(phi) hat(phi)))=-1//r\Gamma^{\hat{\phi}_{\hat{r} \hat{\phi}}}=+1 / r, \Gamma^{\hat{r}_{\hat{\phi} \hat{\phi}}}=-1 / rΓϕ^r^ϕ^=+1/r,Γr^ϕ^ϕ^=1/r; all others vanish.]
(e) Consider the Keplerian orbit of Figure 8.1 and $ 8.3 $ 8.3 $8.3\$ 8.3$8.3 as a nongeodesic curve in the sun's two-dimensional, Euclidean, equatorial plane. In place of the old notation d v / d t , d e γ ^ / d t d v / d t , d e γ ^ / d t dv//dt,de_( hat(gamma))//dtd v / d t, d e_{\hat{\gamma}} / d tdv/dt,deγ^/dt, etc., use the new notation v v , v e r ^ v v , v e r ^ grad_(v)v,grad_(v)e_( hat(r))\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{v}, \boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{e}_{\hat{r}}vv,ver^, etc. Then v = d P / d t v = d P / d t v=dP//dt\boldsymbol{v}=d \mathscr{P} / d tv=dP/dt is the tangent to the orbit, and a = v v a = v v a=grad_(v)v\boldsymbol{a}=\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{v}a=vv is the acceleration. Derive equations (8.4) for a r ^ a r ^ a^( hat(r))a^{\hat{r}}ar^ and a ϕ ^ a ϕ ^ a^( hat(phi))a^{\hat{\phi}}aϕ^ using component manipulations and connection coefficients in the orthonormal basis.

Exercise 8.6. SPHERICAL COORDINATES IN FLAT SPACETIME

The spherical noncoordinate basis { e α ^ } e α ^ {e_( hat(alpha))}\left\{\boldsymbol{e}_{\hat{\alpha}}\right\}{eα^} of Exercise 8.1 was orthonormal, g α ^ β ^ = η α ^ β ^ g α ^ β ^ = η α ^ β ^ g_( hat(alpha) hat(beta))=eta_( hat(alpha) hat(beta))g_{\hat{\alpha} \hat{\beta}}=\eta_{\hat{\alpha} \hat{\beta}}gα^β^=ηα^β^, but had nonvanishing commutation coefficients [part (c) of Exercise 8.2].
(a) Calculate the connection coefficients for this basis, using equations (8.24). [Answer:
Γ θ ^ i θ ^ ^ = Γ ϕ ^ r ^ ϕ ^ = Γ r ^ θ ^ θ ^ = Γ ϕ ^ ϕ ^ r ^ 2 = 1 / r Γ ϕ ^ θ ^ ϕ ^ = Γ θ ^ ϕ ^ ϕ ^ = cot θ Γ θ ^ i θ ^ ^ = Γ ϕ ^ r ^ ϕ ^ = Γ r ^ θ ^ θ ^ = Γ ϕ ^ ϕ ^ r ^ 2 = 1 / r Γ ϕ ^ θ ^ ϕ ^ = Γ θ ^ ϕ ^ ϕ ^ = cot θ {:[Gamma^( hat(theta)) hat(i( hat(theta)))=Gamma^( hat(phi)_( hat(r) hat(phi)))=-Gamma^( hat(r)_( hat(theta) hat(theta)))=-Gamma_( hat(phi) hat(phi))^( hat(r)^(2))=1//r],[Gamma^( hat(phi)_( hat(theta) hat(phi)))=-Gamma^( hat(theta) hat(phi) hat(phi))=cot theta]:}\begin{aligned} \Gamma^{\hat{\theta}} \hat{i \hat{\theta}} & =\Gamma^{\hat{\phi}_{\hat{r} \hat{\phi}}}=-\Gamma^{\hat{r}_{\hat{\theta} \hat{\theta}}}=-\Gamma_{\hat{\phi} \hat{\phi}}^{\hat{r}^{2}}=1 / r \\ \Gamma^{\hat{\phi}_{\hat{\theta} \hat{\phi}}} & =-\Gamma^{\hat{\theta} \hat{\phi} \hat{\phi}}=\cot \theta \end{aligned}Γθ^iθ^^=Γϕ^r^ϕ^=Γr^θ^θ^=Γϕ^ϕ^r^2=1/rΓϕ^θ^ϕ^=Γθ^ϕ^ϕ^=cotθ
all others vanish.]
(b) Write down expressions for α ^ e β ^ α ^ e β ^ grad_( hat(alpha))e_( hat(beta))\boldsymbol{\nabla}_{\hat{\alpha}} \boldsymbol{e}_{\hat{\beta}}α^eβ^ in terms of e γ ^ e γ ^ e_( hat(gamma))\boldsymbol{e}_{\hat{\gamma}}eγ^, and verify the correctness of these expressions by drawing sketches of the basis vectors on a sphere of constant t t ttt and r r rrr. [Answer:
θ ^ e r ^ = r 1 e θ ^ , θ ^ e θ ^ = r 1 e r ^ , ρ ^ e r ^ = r 1 e ϕ ^ δ ^ e θ ^ = ( cot θ / r ) e ϕ ^ , ϕ ^ e ϕ ^ = r 1 e r ^ ( cot θ / r ) e θ ^ θ ^ e r ^ = r 1 e θ ^ , θ ^ e θ ^ = r 1 e r ^ , ρ ^ e r ^ = r 1 e ϕ ^ δ ^ e θ ^ = ( cot θ / r ) e ϕ ^ , ϕ ^ e ϕ ^ = r 1 e r ^ ( cot θ / r ) e θ ^ {:[grad_( hat(theta))e_( hat(r))=r^(-1)e_( hat(theta))","quadgrad_( hat(theta))e_( hat(theta))=-r^(-1)e_( hat(r))","quadgrad_( hat(rho))e_( hat(r))=r^(-1)e_( hat(phi))],[grad_( hat(delta))e_( hat(theta))=(cot theta//r)e_( hat(phi))","quadgrad_( hat(phi))e_( hat(phi))=-r^(-1)e_( hat(r))-(cot theta//r)e_( hat(theta))]:}\begin{aligned} & \boldsymbol{\nabla}_{\hat{\theta}} \boldsymbol{e}_{\hat{r}}=r^{-1} \boldsymbol{e}_{\hat{\theta}}, \quad \boldsymbol{\nabla}_{\hat{\theta}} \boldsymbol{e}_{\hat{\theta}}=-r^{-1} \boldsymbol{e}_{\hat{r}}, \quad \boldsymbol{\nabla}_{\hat{\rho}} \boldsymbol{e}_{\hat{r}}=r^{-1} \boldsymbol{e}_{\hat{\phi}} \\ & \boldsymbol{\nabla}_{\hat{\delta}} \boldsymbol{e}_{\hat{\theta}}=(\cot \theta / r) \boldsymbol{e}_{\hat{\phi}}, \quad \boldsymbol{\nabla}_{\hat{\phi}} \boldsymbol{e}_{\hat{\phi}}=-r^{-1} \boldsymbol{e}_{\hat{r}}-(\cot \theta / r) \boldsymbol{e}_{\hat{\theta}} \end{aligned}θ^er^=r1eθ^,θ^eθ^=r1er^,ρ^er^=r1eϕ^δ^eθ^=(cotθ/r)eϕ^,ϕ^eϕ^=r1er^(cotθ/r)eθ^
All others vanish.]
(c) Calculate the divergence of a vector, A = A α ^ ; α ^ A = A α ^ ; α ^ grad*A=A^( hat(alpha))_(; hat(alpha))\boldsymbol{\nabla} \cdot \boldsymbol{A}=A^{\hat{\alpha}}{ }_{; \hat{\alpha}}A=Aα^;α^, in this basis. [Answer:
A = A t ^ , i ^ + r 2 ( r 2 A r ^ ) , r ^ + ( sin θ ) 1 ( sin θ A θ ^ ) , θ ^ + A ϕ ^ , ϕ ^ = A t ^ t + 1 r 2 ( r 2 A r ^ ) r + 1 r sin θ ( sin θ A θ ^ ) θ + 1 r sin θ A ϕ ^ ϕ . A = A t ^ , i ^ + r 2 r 2 A r ^ , r ^ + ( sin θ ) 1 sin θ A θ ^ , θ ^ + A ϕ ^ , ϕ ^ = A t ^ t + 1 r 2 r 2 A r ^ r + 1 r sin θ sin θ A θ ^ θ + 1 r sin θ A ϕ ^ ϕ . {:[grad*A=A^( hat(t))"," hat(i)+r^(-2)(r^(2)A^( hat(r)))_(, hat(r))+(sin theta)^(-1)(sin thetaA^( hat(theta)))_(, hat(theta))+A^( hat(phi))"," hat(phi)],[=(delA^( hat(t)))/(del t)+(1)/(r^(2))(del(r^(2)A^( hat(r))))/(del r)+(1)/(r sin theta)(del(sin thetaA^( hat(theta))))/(del theta)+(1)/(r sin theta)(delA^( hat(phi)))/(del phi).]:}\begin{aligned} \boldsymbol{\nabla} \cdot \boldsymbol{A} & =A^{\hat{t}}, \hat{i}+r^{-2}\left(r^{2} A^{\hat{r}}\right)_{, \hat{r}}+(\sin \theta)^{-1}\left(\sin \theta A^{\hat{\theta}}\right)_{, \hat{\theta}}+A^{\hat{\phi}}, \hat{\phi} \\ & =\frac{\partial A^{\hat{t}}}{\partial t}+\frac{1}{r^{2}} \frac{\partial\left(r^{2} A^{\hat{r}}\right)}{\partial r}+\frac{1}{r \sin \theta} \frac{\partial\left(\sin \theta A^{\hat{\theta}}\right)}{\partial \theta}+\frac{1}{r \sin \theta} \frac{\partial A^{\hat{\phi}}}{\partial \phi} . \end{aligned}A=At^,i^+r2(r2Ar^),r^+(sinθ)1(sinθAθ^),θ^+Aϕ^,ϕ^=At^t+1r2(r2Ar^)r+1rsinθ(sinθAθ^)θ+1rsinθAϕ^ϕ.
This answer should be familiar from flat-space vector analysis.]

Exercise 8.7. SYMMETRIES OF CONNECTION COEFFICIENTS

From equation (8.24b), the symmetry of the metric, and the antisymmetry ( c β γ μ = c γ β μ c β γ μ = c γ β μ c_(beta gamma mu)=-c_(gamma beta mu)c_{\beta \gamma \mu}=-c_{\gamma \beta \mu}cβγμ=cγβμ )
of the commutation coefficients, show that: Γ α [ β γ ] = 0 Γ α [ β γ ] = 0 Gamma_(alpha[beta gamma])=0\Gamma_{\alpha[\beta \gamma]}=0Γα[βγ]=0 (last two indices are symmetric) in a coordinate basis; Γ ( α ^ β ) γ ^ = 0 Γ ( α ^ β ) γ ^ = 0 Gamma_(( hat(alpha)beta) hat(gamma))=0\Gamma_{(\hat{\alpha} \beta) \hat{\gamma}}=0Γ(α^β)γ^=0 (first two indices are antisymmetric) in a globally orthonormal basis, g α ^ β = η α β g α ^ β = η α β g_( hat(alpha)beta)=eta_(alpha beta)g_{\hat{\alpha} \beta}=\eta_{\alpha \beta}gα^β=ηαβ.

SYSTEMATIC DERIVATION OF RESULTS IN §8.5

Exercise 8.8. NEW DEFINITION OF T T grad T\boldsymbol{\nabla} TT COMPARED WITH OLD DEFINITION

The new definition of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T is given by equations (8.16) and (8.17). Use the fact that parallel transport keeps local-Lorentz components fixed to derive, from (8.16), the Lorentz-frame equation u T = T β α , γ u γ e β ω α u T = T β α , γ u γ e β ω α grad_(u)T=T^(beta)_(alpha,gamma)u^(gamma)e_(beta)oxomega^(alpha)\boldsymbol{\nabla}_{u} \boldsymbol{T}=T^{\beta}{ }_{\alpha, \gamma} u^{\gamma} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}uT=Tβα,γuγeβωα. From this and equation (8.17), infer that the Lorentzframe components of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T are T β α , γ T β α , γ T^(beta)_(alpha,gamma)T^{\beta}{ }_{\alpha, \gamma}Tβα,γ-which accords with the old definition of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T.

Exercise 8.9. CHAIN RULE FOR u T u T grad_(u)T\nabla_{u} TuT

(a) Use calculations in a local Lorentz frame to show that " u u grad_(u)\boldsymbol{\nabla}_{u}u " obeys the standard chain rule for derivatives:
(8.27) u ( f A B ) = ( u f ) A B + f ( u A ) B + f A ( u B ) (8.27) u ( f A B ) = u f A B + f u A B + f A u B {:(8.27)grad_(u)(fA ox B)=(grad_(u)f)A ox B+f(grad_(u)A)ox B+fA ox(grad_(u)B):}\begin{equation*} \boldsymbol{\nabla}_{u}(f \boldsymbol{A} \otimes \boldsymbol{B})=\left(\boldsymbol{\nabla}_{u} f\right) \boldsymbol{A} \otimes \boldsymbol{B}+f\left(\boldsymbol{\nabla}_{u} \boldsymbol{A}\right) \otimes \boldsymbol{B}+f \boldsymbol{A} \otimes\left(\boldsymbol{\nabla}_{u} \boldsymbol{B}\right) \tag{8.27} \end{equation*}(8.27)u(fAB)=(uf)AB+f(uA)B+fA(uB)
Here A A A\boldsymbol{A}A and B B B\boldsymbol{B}B are arbitrary vectors, 1 -forms, or tensors; and f f fff is an arbitrary function. [Hint: assume for concreteness that A A A\boldsymbol{A}A is a ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11) tensor and B B B\boldsymbol{B}B is a vector. Then this equation reads, in Lorentz-frame component notation,
( ) ( f A α β B γ ) , δ u δ = ( f , δ u δ ) A α β B γ + f ( A α β , δ u δ ) B γ + f A α β ( B γ , δ u δ ) ] ( ) f A α β B γ , δ u δ = f , δ u δ A α β B γ + f A α β , δ u δ B γ + f A α β B γ , δ u δ {:('")"{:(fA^(alpha)_(beta)B^(gamma))_(,delta)u^(delta)=(f_(,delta)u^(delta))A^(alpha)_(beta)B^(gamma)+f(A^(alpha)_(beta,delta)u^(delta))B^(gamma)+fA^(alpha)_(beta)(B^(gamma)_(,delta)u^(delta))*]:}\begin{equation*} \left.\left(f A^{\alpha}{ }_{\beta} B^{\gamma}\right)_{, \delta} u^{\delta}=\left(f_{, \delta} u^{\delta}\right) A^{\alpha}{ }_{\beta} B^{\gamma}+f\left(A^{\alpha}{ }_{\beta, \delta} u^{\delta}\right) B^{\gamma}+f A^{\alpha}{ }_{\beta}\left(B^{\gamma}{ }_{, \delta} u^{\delta}\right) \cdot\right] \tag{$\prime$} \end{equation*}()(fAαβBγ),δuδ=(f,δuδ)AαβBγ+f(Aαβ,δuδ)Bγ+fAαβ(Bγ,δuδ)]
(b) Rewrite equation (8.27) in component notation in an arbitrary basis. [Answer: same as (8.27'), except "," is replaced everywhere by ";". But note that f ; δ u δ = f δ δ u δ f ; δ u δ = f δ δ u δ f_(;delta)u^(delta)=f_(delta delta)u^(delta)f_{; \delta} u^{\delta}=f_{\delta \delta} u^{\delta}f;δuδ=fδδuδ, because the function f f fff "has no components to correct".]

Exercise 8.10. COVARIANT DERIVATIVE COMMUTES WITH CONTRACTION

(a) Let S S S\boldsymbol{S}S be a ( 1 2 ) 1 2 ((1)/(2))\left(\frac{1}{2}\right)(12) tensor. Using components in a local Lorentz frame show that
(8.28) u ( contraction on slots 1 and 2 of S ) = ( contraction on slots 1 and 2 of u S ) (8.28) u (  contraction on slots  1  and  2  of  S ) =  contraction on slots  1  and  2  of  u S {:(8.28)grad_(u)(" contraction on slots "1" and "2" of "S)=(" contraction on slots "1" and "2" of "grad_(u)S):}\begin{equation*} \boldsymbol{\nabla}_{u}(\text { contraction on slots } 1 \text { and } 2 \text { of } \boldsymbol{S})=\left(\text { contraction on slots } 1 \text { and } 2 \text { of } \boldsymbol{\nabla}_{u} \boldsymbol{S}\right) \tag{8.28} \end{equation*}(8.28)u( contraction on slots 1 and 2 of S)=( contraction on slots 1 and 2 of uS)
[Hint: in a local Lorentz frame this equation makes the trivial statement
( α S α β α ) , γ u γ = α ( S α β , γ α u γ ) ] α S α β α , γ u γ = α S α β , γ α u γ {:(sum_(alpha)S_(alpha beta)^(alpha))_(,gamma)u^(gamma)=sum_(alpha)(S_(alpha beta,gamma)^(alpha)u^(gamma))*]\left.\left(\sum_{\alpha} S_{\alpha \beta}^{\alpha}\right)_{, \gamma} u^{\gamma}=\sum_{\alpha}\left(S_{\alpha \beta, \gamma}^{\alpha} u^{\gamma}\right) \cdot\right](αSαβα),γuγ=α(Sαβ,γαuγ)]

Exercise 8.11. ALGEBRAIC PROPERTIES OF grad\boldsymbol{\nabla}

Use calculations in a local Lorentz frame to show that
(8.29) a u + b v s = a u s + b v s (8.29) a u + b v s = a u s + b v s {:(8.29)grad_(au+bv)s=agrad_(u)s+bgrad_(v)s:}\begin{equation*} \boldsymbol{\nabla}_{a u+b v} \boldsymbol{s}=a \boldsymbol{\nabla}_{u} \boldsymbol{s}+b \boldsymbol{\nabla}_{\mathbf{v}} \boldsymbol{s} \tag{8.29} \end{equation*}(8.29)au+bvs=aus+bvs
for all tangent vectors u , v u , v u,v\boldsymbol{u}, \boldsymbol{v}u,v and numbers a , b a , b a,ba, ba,b; also that
(8.30) u ( S + M ) = u S + u M (8.30) u ( S + M ) = u S + u M {:(8.30)grad_(u)(S+M)=grad_(u)S+grad_(u)M:}\begin{equation*} \nabla_{u}(S+\boldsymbol{M})=\nabla_{u} S+\nabla_{u} \boldsymbol{M} \tag{8.30} \end{equation*}(8.30)u(S+M)=uS+uM
for any two tensor fields S S S\boldsymbol{S}S and M M M\boldsymbol{M}M of the same rank; also that

for any two vector fields u u u\boldsymbol{u}u and w w w\boldsymbol{w}w.

Exercise 8.12. CONNECTION COEFFICIENTS FOR 1-FORM BASIS

Show that the same connection coefficients Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ that describe the changes in { e β } e β {e_(beta)}\left\{\boldsymbol{e}_{\beta}\right\}{eβ} from point to point [definition (8.19a)] also describe the changes in { ω α } ω α {omega^(alpha)}\left\{\boldsymbol{\omega}^{\alpha}\right\}{ωα}, except for a change in sign [equation (8.19b)]. { {:}\left\{\right.{ Answer: (1) ω α , e β = δ α β ω α , e β = δ α β (:omega^(alpha),e_(beta):)=delta^(alpha)_(beta)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=\delta^{\alpha}{ }_{\beta}ωα,eβ=δαβ is a constant function ( 0 or 1 , depending on whether α = β α = β alpha=beta\alpha=\betaα=β ). (2) Thus, γ ω α , e β = e γ ω α , e β = 0 γ ω α , e β = e γ ω α , e β = 0 grad_(gamma)(:omega^(alpha),e_(beta):)=del_(e_(gamma))(:omega^(alpha),e_(beta):)=0\nabla_{\gamma}\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=\partial_{\boldsymbol{e}_{\gamma}}\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=0γωα,eβ=eγωα,eβ=0. (3) But ω α , e β ω α , e β (:omega^(alpha),e_(beta):)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangleωα,eβ is the contraction of ω α e β ω α e β omega^(alpha)oxe_(beta)\boldsymbol{\omega}^{\alpha} \otimes \boldsymbol{e}_{\beta}ωαeβ, so equation (8.28) implies 0 = γ ( 0 = γ 0=grad_(gamma)(:}0=\boldsymbol{\nabla}_{\gamma}\left(\right.0=γ( contraction of ω α e β ω α e β omega^(alpha)oxe_(beta)\boldsymbol{\omega}^{\alpha} \otimes \boldsymbol{e}_{\beta}ωαeβ ) = = === contraction of [ γ ( ω α e β ) ] γ ω α e β [grad gamma(omega^(alpha)oxe_(beta))]\left[\boldsymbol{\nabla} \gamma\left(\boldsymbol{\omega}^{\alpha} \otimes \boldsymbol{e}_{\beta}\right)\right][γ(ωαeβ)]. (4) Apply the chain rule (8.27) to conclude 0 = 0 = 0=0=0= contraction of [ ( γ ω α ) e β + ω α ( γ e β ) ] = γ ω α , e β + ω α , γ e β γ ω α e β + ω α γ e β = γ ω α , e β + ω α , γ e β [(grad_(gamma)omega^(alpha))oxe_(beta)+omega^(alpha)ox(grad_(gamma)e_(beta))]=(:grad_(gamma)omega^(alpha),e_(beta):)+(:omega^(alpha),grad_(gamma)e_(beta):)\left[\left(\boldsymbol{\nabla}_{\gamma} \boldsymbol{\omega}^{\alpha}\right) \otimes \boldsymbol{e}_{\beta}+\boldsymbol{\omega}^{\alpha} \otimes\left(\boldsymbol{\nabla}_{\gamma} \boldsymbol{e}_{\beta}\right)\right]=\left\langle\boldsymbol{\nabla}_{\gamma} \boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle+\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{\nabla}_{\gamma} \boldsymbol{e}_{\beta}\right\rangle[(γωα)eβ+ωα(γeβ)]=γωα,eβ+ωα,γeβ. (5) Finally, use definition (8.19a) to arrive at the desired result, (8.19b).}

Exercise 8.13. " Γ T Γ T GammaT^('')\Gamma T^{\prime \prime}ΓT CORRECTION TERMS FOR T β a ; γ T β a ; γ T^(beta)_(a;gamma)T^{\beta}{ }_{a ; \gamma}Tβa;γ

Derive equation (8.21) for T β α ; γ T β α ; γ T^(beta)_(alpha;gamma)T^{\beta}{ }_{\alpha ; \gamma}Tβα;γ in an arbitrary basis by first calculating the components of u T u T grad_(u)T\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}uT for arbitrary u u u\boldsymbol{u}u, and by then using equation (8.17) to infer the components of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T. [Answer: (1) Use the chain rule (8.27) to get
u T = u ( T β α e β ω α ) = ( u T β α ) e β ω α + T β α ( u e β ) w α + T β α e β ( u ω α ) . u T = u T β α e β ω α = u T β α e β ω α + T β α u e β w α + T β α e β u ω α . {:[grad_(u)T=grad_(u)(T^(beta)_(alpha)e_(beta)oxomega^(alpha))],[=(grad_(u)T^(beta)_(alpha))e_(beta)oxomega^(alpha)+T^(beta)_(alpha)(grad_(u)e_(beta))oxw^(alpha)+T^(beta)_(alpha)e_(beta)ox(grad_(u)omega^(alpha)).]:}\begin{aligned} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T} & =\boldsymbol{\nabla}_{\boldsymbol{u}}\left(T^{\beta}{ }_{\alpha} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}\right) \\ & =\left(\boldsymbol{\nabla}_{\boldsymbol{u}} T^{\beta}{ }_{\alpha}\right) \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}+T^{\beta}{ }_{\alpha}\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{e}_{\beta}\right) \otimes \boldsymbol{w}^{\alpha}+T^{\beta}{ }_{\alpha} \boldsymbol{e}_{\beta} \otimes\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\omega}^{\alpha}\right) . \end{aligned}uT=u(Tβαeβωα)=(uTβα)eβωα+Tβα(ueβ)wα+Tβαeβ(uωα).
(2) Write u u u\boldsymbol{u}u in terms of its components, u = u γ e γ u = u γ e γ u=u^(gamma)e_(gamma)\boldsymbol{u}=u^{\gamma} \boldsymbol{e}_{\gamma}u=uγeγ; use linearity of u u grad_(u)\boldsymbol{\nabla}_{\boldsymbol{u}}u in u u u\boldsymbol{u}u from equation (8.29), to get u = u γ γ u = u γ γ grad_(u)=u^(gamma)grad_(gamma)\boldsymbol{\nabla}_{u}=u^{\gamma} \boldsymbol{\nabla}_{\gamma}u=uγγ; and use this in u T u T grad_(u)T\boldsymbol{\nabla}_{u} \boldsymbol{T}uT :
u T = u γ { T α , γ β e β ω α + T α β ( γ e β ) ω α + T β α e β ( γ ω α ) } u T = u γ T α , γ β e β ω α + T α β γ e β ω α + T β α e β γ ω α grad_(u)T=u^(gamma){T_(alpha,gamma)^(beta)e_(beta)oxomega^(alpha)+T_(alpha)^(beta)(grad_(gamma)e_(beta))oxomega^(alpha)+T^(beta)_(alpha)e_(beta)ox(grad_(gamma)omega^(alpha))}\boldsymbol{\nabla}_{u} \boldsymbol{T}=u^{\gamma}\left\{T_{\alpha, \gamma}^{\beta} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}+T_{\alpha}^{\beta}\left(\boldsymbol{\nabla}_{\gamma} \boldsymbol{e}_{\beta}\right) \otimes \boldsymbol{\omega}^{\alpha}+T^{\beta}{ }_{\alpha} \boldsymbol{e}_{\beta} \otimes\left(\boldsymbol{\nabla}_{\gamma} \boldsymbol{\omega}^{\alpha}\right)\right\}uT=uγ{Tα,γβeβωα+Tαβ(γeβ)ωα+Tβαeβ(γωα)}
(3) Use equations ( 8.19 a , b ) ( 8.19 a , b ) (8.19 a,b)(8.19 a, b)(8.19a,b), rewritten as
(8.32) γ e β = Γ β γ μ e μ , γ ω α = Γ α μ γ ω μ , (8.32) γ e β = Γ β γ μ e μ , γ ω α = Γ α μ γ ω μ , {:(8.32)grad_(gamma)e_(beta)=Gamma_(beta gamma)^(mu)e_(mu)","quadgrad_(gamma)omega^(alpha)=-Gamma^(alpha)_(mu gamma)omega^(mu)",":}\begin{equation*} \boldsymbol{\nabla}_{\gamma} \boldsymbol{e}_{\beta}=\Gamma_{\beta \gamma}^{\mu} \boldsymbol{e}_{\mu}, \quad \boldsymbol{\nabla}_{\gamma} \boldsymbol{\omega}^{\alpha}=-\Gamma^{\alpha}{ }_{\mu \gamma} \boldsymbol{\omega}^{\mu}, \tag{8.32} \end{equation*}(8.32)γeβ=Γβγμeμ,γωα=Γαμγωμ,
to put u T u T grad_(u)T\boldsymbol{\nabla}_{u} \boldsymbol{T}uT in the form
u T = u γ { T β α , γ e β ω α + Γ μ β γ T β α e μ ω α Γ α μ γ T β α e β ω μ } . u T = u γ T β α , γ e β ω α + Γ μ β γ T β α e μ ω α Γ α μ γ T β α e β ω μ . grad_(u)T=u^(gamma){T^(beta)_(alpha,gamma)e_(beta)oxomega^(alpha)+Gamma^(mu)_(beta gamma)T^(beta)_(alpha)e_(mu)oxomega^(alpha)-Gamma^(alpha)_(mu gamma)T^(beta)_(alpha)e_(beta)oxomega^(mu)}.\boldsymbol{\nabla}_{u} \boldsymbol{T}=u^{\gamma}\left\{T^{\beta}{ }_{\alpha, \gamma} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}+\Gamma^{\mu}{ }_{\beta \gamma} T^{\beta}{ }_{\alpha} \boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\alpha}-\Gamma^{\alpha}{ }_{\mu \gamma} T^{\beta}{ }_{\alpha} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\mu}\right\} .uT=uγ{Tβα,γeβωα+ΓμβγTβαeμωαΓαμγTβαeβωμ}.
(4) Rename dummy indices so that the basis tensor e β ω α e β ω α e_(beta)oxomega^(alpha)\boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha}eβωα can be factored out:
u T = u γ { T β α , γ + Γ β μ γ T μ α Γ α γ μ T β μ } e β ω α . u T = u γ T β α , γ + Γ β μ γ T μ α Γ α γ μ T β μ e β ω α . grad_(u)T=u^(gamma){T^(beta)_(alpha,gamma)+Gamma^(beta)_(mu gamma)T^(mu)_(alpha)-Gamma_(alpha gamma)^(mu)T^(beta)_(mu)}e_(beta)oxomega^(alpha).\boldsymbol{\nabla}_{u} \boldsymbol{T}=u^{\gamma}\left\{T^{\beta}{ }_{\alpha, \gamma}+\Gamma^{\beta}{ }_{\mu \gamma} T^{\mu}{ }_{\alpha}-\Gamma_{\alpha \gamma}^{\mu} T^{\beta}{ }_{\mu}\right\} \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha} .uT=uγ{Tβα,γ+ΓβμγTμαΓαγμTβμ}eβωα.
(5) By comparison with
u T = T ( , , u ) = ( T β α ; γ u γ ) e β ω α , u T = T ( , , u ) = T β α ; γ u γ e β ω α , grad_(u)T=grad T(dots,dots,u)=(T^(beta)_(alpha;gamma)u^(gamma))e_(beta)oxomega^(alpha),\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{T}=\boldsymbol{\nabla} \boldsymbol{T}(\ldots, \ldots, \boldsymbol{u})=\left(T^{\beta}{ }_{\alpha ; \gamma} u^{\gamma}\right) \boldsymbol{e}_{\beta} \otimes \boldsymbol{\omega}^{\alpha},uT=T(,,u)=(Tβα;γuγ)eβωα,
read off the value of T β α ; γ ] ] T β α ; γ ] {:T^(beta)_(alpha;gamma)^(])]\left.T^{\beta}{ }_{\alpha ; \gamma}{ }^{]}\right]Tβα;γ]]

Exercise 8.14. METRIC IS COVARIANTLY CONSTANT

Show on physical grounds (using properties of local Lorentz frames) that
(8.33) g = 0 (8.33) g = 0 {:(8.33)grad g=0:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{g}=0 \tag{8.33} \end{equation*}(8.33)g=0
or, equivalently, that u g = 0 u g = 0 grad_(u)g=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{g}=0ug=0 for any vector u u u\boldsymbol{u}u. Then deduce as a mathematical consequence the obviously desirable product rule
u ( A B ) = ( u A ) B + A ( u B ) u ( A B ) = u A B + A u B grad_(u)(A*B)=(grad_(u)A)*B+A*(grad_(u)B)\nabla_{u}(\boldsymbol{A} \cdot \boldsymbol{B})=\left(\boldsymbol{\nabla}_{u} \boldsymbol{A}\right) \cdot \boldsymbol{B}+\boldsymbol{A} \cdot\left(\boldsymbol{\nabla}_{u} \boldsymbol{B}\right)u(AB)=(uA)B+A(uB)
[Answer: (1) As discussed following equation (8.18), the components of g g grad g\boldsymbol{\nabla} \boldsymbol{g}g in a local Lorentz frame are g μ ν , α g μ ν , α g_(mu nu,alpha^(**))g_{\mu \nu, \alpha^{*}}gμν,α Just use g g g\boldsymbol{g}g for T T T\boldsymbol{T}T in that discussion. But these components all vanish by equation (8.15b). Therefore equation (8.33) holds in this frame, and-as a tensor equation-in all frames. (2) The product rule is also a tensor equation, true immediately via components in a local Lorentz frame. (3) Prove the product rule also the hard way, to see where equation (8.33) enters. Use the chain rule of exercise 8.9 to write
u ( g A B ) = ( u g ) A B + g ( u A ) B + g A ( u B ) u ( g A B ) = u g A B + g u A B + g A u B {:[grad_(u)(g ox A ox B)=(grad_(u)g)ox A ox B+g ox(grad_(u)A)ox B],[+g ox A ox(grad_(u)B)]:}\begin{aligned} \nabla_{u}(g \otimes A \otimes B)= & \left(\nabla_{u} g\right) \otimes A \otimes B+g \otimes\left(\nabla_{u} A\right) \otimes B \\ & +g \otimes A \otimes\left(\nabla_{u} B\right) \end{aligned}u(gAB)=(ug)AB+g(uA)B+gA(uB)
Use equation (8.33) to drop one term, then contract, forming
A B = contraction ( g A B ) A B = contraction ( g A B ) A*B=contraction(g ox A ox B)\boldsymbol{A} \cdot \boldsymbol{B}=\operatorname{contraction}(\boldsymbol{g} \otimes \boldsymbol{A} \otimes \boldsymbol{B})AB=contraction(gAB)
and the other inner products. Exercise 8.10 is used to justify commuting the contraction with u u grad_(u)\boldsymbol{\nabla}_{u}u on the lefthand side.]

Exercise 8.15. CONNECTION COEFFICIENTS IN TERMS OF METRIC

Use the fact that the metric is covariantly constant [equation (8.33)] to derive equation (8.24b) for the connection coefficients. Treat equation (8.24c) as a definition of Γ μ β γ Γ μ β γ Gamma_(mu beta gamma)\Gamma_{\mu \beta \gamma}Γμβγ in terms of Γ β γ α Γ β γ α Gamma_(betagamma^('))^(alpha)\Gamma_{\beta \gamma^{\prime}}^{\alpha}Γβγα [Answer: (1) Calculate the components of g g grad g\boldsymbol{\nabla} \boldsymbol{g}g in an arbitrary frame:
g α β ; γ = 0 = g α β , γ Γ μ α γ g μ β Γ μ β γ g μ α g α β , γ Γ β α γ Γ α β γ ; g α β ; γ = 0 = g α β , γ Γ μ α γ g μ β Γ μ β γ g μ α g α β , γ Γ β α γ Γ α β γ ; {:[g_(alpha beta;gamma)=0=g_(alpha beta,gamma)-Gamma^(mu)_(alpha gamma)g_(mu beta)-Gamma^(mu)_(beta gamma)g_(mu alpha)],[-=g_(alpha beta,gamma)-Gamma_(beta alpha gamma)-Gamma_(alpha beta gamma);]:}\begin{aligned} g_{\alpha \beta ; \gamma} & =0=g_{\alpha \beta, \gamma}-\Gamma^{\mu}{ }_{\alpha \gamma} g_{\mu \beta}-\Gamma^{\mu}{ }_{\beta \gamma} g_{\mu \alpha} \\ & \equiv g_{\alpha \beta, \gamma}-\Gamma_{\beta \alpha \gamma}-\Gamma_{\alpha \beta \gamma} ; \end{aligned}gαβ;γ=0=gαβ,γΓμαγgμβΓμβγgμαgαβ,γΓβαγΓαβγ;
thereby conclude that g α β , γ = 2 Γ ( α β ) γ g α β , γ = 2 Γ ( α β ) γ g_(alpha beta,gamma)=2Gamma_((alpha beta)gamma)g_{\alpha \beta, \gamma}=2 \Gamma_{(\alpha \beta) \gamma}gαβ,γ=2Γ(αβ)γ. (Round brackets denote symmetric part.) (2) Construct the metric terms in the claimed answer for Γ μ β γ Γ μ β γ Gamma_(mu beta gamma)\Gamma_{\mu \beta \gamma}Γμβγ :
1 2 ( g μ β , γ + g μ γ , β g β γ , μ ) = Γ ( μ β ) γ + Γ ( μ γ ) β Γ ( β γ ) μ = 1 2 [ Γ μ β γ + Γ β μ γ + Γ μ γ β + Γ γ μ β Γ β γ μ Γ γ β μ ] = Γ μ β γ + ( Γ μ [ β γ ] + Γ β [ μ γ ] + Γ γ [ μ β ] ) 1 2 g μ β , γ + g μ γ , β g β γ , μ = Γ ( μ β ) γ + Γ ( μ γ ) β Γ ( β γ ) μ = 1 2 Γ μ β γ + Γ β μ γ + Γ μ γ β + Γ γ μ β Γ β γ μ Γ γ β μ = Γ μ β γ + Γ μ [ β γ ] + Γ β [ μ γ ] + Γ γ [ μ β ] {:[(1)/(2)(g_(mu beta,gamma)+g_(mu gamma,beta)-g_(beta gamma,mu))=Gamma_((mu beta)gamma)+Gamma_((mu gamma)beta)-Gamma_((beta gamma)mu)],[=(1)/(2)[Gamma_(mu beta gamma)+Gamma_(beta mu gamma)+Gamma_(mu gamma beta)+Gamma_(gamma mu beta)-Gamma_(beta gamma mu)-Gamma_(gamma beta mu)]],[=Gamma_(mu beta gamma)+(-Gamma_(mu[beta gamma])+Gamma_(beta[mu gamma])+Gamma_(gamma[mu beta]))]:}\begin{aligned} \frac{1}{2}\left(g_{\mu \beta, \gamma}+g_{\mu \gamma, \beta}-g_{\beta \gamma, \mu}\right) & =\Gamma_{(\mu \beta) \gamma}+\Gamma_{(\mu \gamma) \beta}-\Gamma_{(\beta \gamma) \mu} \\ & =\frac{1}{2}\left[\Gamma_{\mu \beta \gamma}+\Gamma_{\beta \mu \gamma}+\Gamma_{\mu \gamma \beta}+\Gamma_{\gamma \mu \beta}-\Gamma_{\beta \gamma \mu}-\Gamma_{\gamma \beta \mu}\right] \\ & =\Gamma_{\mu \beta \gamma}+\left(-\Gamma_{\mu[\beta \gamma]}+\Gamma_{\beta[\mu \gamma]}+\Gamma_{\gamma[\mu \beta]}\right) \end{aligned}12(gμβ,γ+gμγ,βgβγ,μ)=Γ(μβ)γ+Γ(μγ)βΓ(βγ)μ=12[Γμβγ+Γβμγ+Γμγβ+ΓγμβΓβγμΓγβμ]=Γμβγ+(Γμ[βγ]+Γβ[μγ]+Γγ[μβ])
(3) Infer from equation (8.31), with u u u\boldsymbol{u}u and w w w\boldsymbol{w}w chosen as two basis vectors ( u = e μ , w = e p u = e μ , w = e p u=e_(mu),w=e_(p)\boldsymbol{u}=\boldsymbol{e}_{\mu}, \boldsymbol{w}=\boldsymbol{e}_{p}u=eμ,w=ep ) that
c μ ν ρ e ρ [ e μ , e ν ] = μ e ν ν e μ = ( Γ ρ ν μ Γ ρ μ ν ) e ρ = 2 Γ [ ν μ ] ρ e ρ ; c μ ν ρ e ρ e μ , e ν = μ e ν ν e μ = Γ ρ ν μ Γ ρ μ ν e ρ = 2 Γ [ ν μ ] ρ e ρ ; c_(mu nu)^(rho)e_(rho)-=[e_(mu),e_(nu)]=grad_(mu)e_(nu)-grad_(nu)e_(mu)=(Gamma^(rho)_(nu mu)-Gamma^(rho)_(mu nu))e_(rho)=2Gamma_([nu mu])^(rho)e_(rho);c_{\mu \nu}{ }^{\rho} \boldsymbol{e}_{\rho} \equiv\left[\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right]=\boldsymbol{\nabla}_{\mu} \boldsymbol{e}_{\nu}-\nabla_{\nu} \boldsymbol{e}_{\mu}=\left(\Gamma^{\rho}{ }_{\nu \mu}-\Gamma^{\rho}{ }_{\mu \nu}\right) \boldsymbol{e}_{\rho}=2 \Gamma_{[\nu \mu]}^{\rho} \boldsymbol{e}_{\rho} ;cμνρeρ[eμ,eν]=μeννeμ=(ΓρνμΓρμν)eρ=2Γ[νμ]ρeρ;
i.e.,
(8.34) Γ [ μ ν ] ρ = 1 2 c μ ν ρ ; Γ ρ [ μ ν ] = 1 2 c μ ν ρ . (8.34) Γ [ μ ν ] ρ = 1 2 c μ ν ρ ; Γ ρ [ μ ν ] = 1 2 c μ ν ρ . {:(8.34)Gamma_([mu nu])^(rho)=-(1)/(2)c_(mu nu)^(rho);quadGamma_(rho[mu nu])=-(1)/(2)c_(mu nu rho).:}\begin{equation*} \Gamma_{[\mu \nu]}^{\rho}=-\frac{1}{2} c_{\mu \nu}^{\rho} ; \quad \Gamma_{\rho[\mu \nu]}=-\frac{1}{2} c_{\mu \nu \rho} . \tag{8.34} \end{equation*}(8.34)Γ[μν]ρ=12cμνρ;Γρ[μν]=12cμνρ.
(4) This, combined with step (2) yields the desired formula for Γ μ β γ Γ μ β γ Gamma_(mu beta gamma)\Gamma_{\mu \beta \gamma}Γμβγ ]

§8.6. LOCAL LORENTZ FRAMES: MATHEMATICAL DISCUSSION

An observer falling freely in curved spacetime makes measurements in his local Lorentz frame. What he discovers has been discussed extensively in Parts I and II of this book. Try now to derive his basic discoveries from the formalism of the last section.
Pick an event P o P o P_(o)\mathscr{P}_{o}Po on the observer's world line. His local Lorentz frame there is a coordinate system x α ( P ) x α ( P ) x^(alpha)(P)x^{\alpha}(\mathscr{P})xα(P) in which
(8.35a) g α β e α e β P x α P x β = η α β at P o (8.35a) g α β e α e β P x α P x β = η α β  at  P o {:(8.35a)g_(alpha beta)-=e_(alpha)*e_(beta)-=(delP)/(delx^(alpha))*(delP)/(delx^(beta))=eta_(alpha beta)" at "P_(o):}\begin{equation*} g_{\alpha \beta} \equiv \boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta} \equiv \frac{\partial \mathscr{P}}{\partial x^{\alpha}} \cdot \frac{\partial \mathscr{P}}{\partial x^{\beta}}=\eta_{\alpha \beta} \text { at } \mathscr{P}_{o} \tag{8.35a} \end{equation*}(8.35a)gαβeαeβPxαPxβ=ηαβ at Po
(Lorentz metric at P o P o P_(o)\mathscr{P}_{o}Po ), and in which
(8.35b) g α β / x μ = 0 at P o (8.35b) g α β / x μ = 0  at  P o {:(8.35b)delg_(alpha beta)//delx^(mu)=0" at "P_(o):}\begin{equation*} \partial g_{\alpha \beta} / \partial x^{\mu}=0 \text { at } \mathscr{P}_{o} \tag{8.35b} \end{equation*}(8.35b)gαβ/xμ=0 at Po
(metric as Lorentz as possible near P o P o P_(o)\mathscr{P}_{o}Po ). [See equation (8.15).] In addition, by virtue of equations (8.24),
(8.36) Γ β γ α = 0 at P o (8.36) Γ β γ α = 0  at  P o {:(8.36)Gamma_(beta gamma)^(alpha)=0" at "P_(o):}\begin{equation*} \Gamma_{\beta \gamma}^{\alpha}=0 \text { at } \mathscr{P}_{o} \tag{8.36} \end{equation*}(8.36)Γβγα=0 at Po
(no "correction terms" in covariant derivatives). Of course, the observer must be at rest in his local Lorentz frame; i.e., his world line must be
(8.37) x j = x j ( P o ) = constant ; x 0 varying. (8.37) x j = x j P o =  constant  ; x 0  varying.  {:(8.37)x^(j)=x^(j)(P_(o))=" constant ";quadx^(0)" varying. ":}\begin{equation*} x^{j}=x^{j}\left(\mathscr{P}_{o}\right)=\text { constant } ; \quad x^{0} \text { varying. } \tag{8.37} \end{equation*}(8.37)xj=xj(Po)= constant ;x0 varying. 
Query: Equations (8.35) to (8.37) guarantee that the observer is at rest in a local Lorentz frame. Do they imply that he is freely falling? (They should!) Answer: Calculate the observer's 4-acceleration a = d u / d τ a = d u / d τ a=du//d tau\boldsymbol{a}=d \boldsymbol{u} / d \taua=du/dτ (notation of chapter 6 ) = u u = u u =grad_(u)u=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=uu (notation of this chapter). His 4 -velocity, calculated from equation (8.37) is
(8.38) u = ( d x α / d τ ) e α = ( d x 0 / d τ ) e 0 = e 0 ; [ because u and e 0 both have unit length ] (8.38) u = d x α / d τ e α = d x 0 / d τ e 0 = e 0 ;  because  u  and  e 0  both   have unit length  {:[(8.38)u=(dx^(alpha)//d tau)e_(alpha)=(dx^(0)//d tau)e_(0)=e_(0);],[[[" because "u" and "e_(0)" both "],[" have unit length "]]]:}\begin{align*} \boldsymbol{u}= & \left(d x^{\alpha} / d \tau\right) \boldsymbol{e}_{\alpha}=\left(d x^{0} / d \tau\right) \boldsymbol{e}_{0}=\boldsymbol{e}_{0} ; \tag{8.38}\\ & {\left[\begin{array}{c} \text { because } \boldsymbol{u} \text { and } \boldsymbol{e}_{0} \text { both } \\ \text { have unit length } \end{array}\right] } \end{align*}(8.38)u=(dxα/dτ)eα=(dx0/dτ)e0=e0;[ because u and e0 both  have unit length ]
so his 4-acceleration is
a = u u = 0 e 0 = Γ α 00 e α (8.39) = 0 at P 0 . a = u u = 0 e 0 = Γ α 00 e α (8.39) = 0  at  P 0 . {:[a=grad_(u)u=grad_(0)e_(0)=Gamma^(alpha)_(00)e_(alpha)],[(8.39)=0" at "P_(0).]:}\begin{align*} \boldsymbol{a}=\boldsymbol{\nabla}_{u} \boldsymbol{u}=\boldsymbol{\nabla}_{0} \boldsymbol{e}_{0} & =\Gamma^{\alpha}{ }_{00} \boldsymbol{e}_{\alpha} \\ & =0 \text { at } \mathscr{P}_{0} . \tag{8.39} \end{align*}a=uu=0e0=Γα00eα(8.39)=0 at P0.
Thus, he is indeed freely falling ( a = 0 ) ( a = 0 ) (a=0)(\boldsymbol{a}=0)(a=0); and he moves along a geodesic ( u u = 0 ) u u = 0 (grad_(u)u=0)\left(\boldsymbol{\nabla}_{u} \boldsymbol{u}=0\right)(uu=0).
Query: Do freely falling particles move along straight lines ( d 2 x α / d τ 2 = 0 ) d 2 x α / d τ 2 = 0 (d^(2)x^(alpha)//dtau^(2)=0)\left(d^{2} x^{\alpha} / d \tau^{2}=0\right)(d2xα/dτ2=0) in the observer's local Lorentz frame at P o P o P_(o)\mathscr{P}_{o}Po ? (They should!) Answer: A freely falling particle experiences zero 4-acceleration
a particle = u particle u particle = 0 a particle  = u particle  u particle  = 0 a_("particle ")=grad_(u_("particle "))u_("particle ")=0\boldsymbol{a}_{\text {particle }}=\boldsymbol{\nabla}_{\boldsymbol{u}_{\text {particle }}} \boldsymbol{u}_{\text {particle }}=0aparticle =uparticle uparticle =0
i.e., it parallel-transports its 4-velocity; i.e., it moves along a geodesic of spacetime
Origin falls freely along a geodesic
with affine parameter equal to its proper time. The geodesic equation for its world line, in local Lorentz coordinates, says
d 2 x α d τ 2 = Γ α μ ν d x μ d τ d x ν d τ = 0 at P o . d 2 x α d τ 2 = Γ α μ ν d x μ d τ d x ν d τ = 0  at  P o . {:[(d^(2)x^(alpha))/(dtau^(2))=-Gamma^(alpha)_(mu nu)(dx^(mu))/(d tau)(dx^(nu))/(d tau)],[=0" at "P_(o).]:}\begin{aligned} \frac{d^{2} x^{\alpha}}{d \tau^{2}} & =-\Gamma^{\alpha}{ }_{\mu \nu} \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu}}{d \tau} \\ & =0 \text { at } \mathscr{P}_{o} . \end{aligned}d2xαdτ2=Γαμνdxμdτdxνdτ=0 at Po.
The particle's world line is, indeed, straight at P 0 P 0 P_(0)\mathscr{P}_{0}P0.
Query: Does the freely falling observer Fermi-Walker-transport his spatial basis vectors e j e j e_(j)\boldsymbol{e}_{j}ej; i.e., can he attach them to gyroscopes that he carries? (He should be able to!) Answer: Fermi-Walker transport (Box 6.2) would say
But u = e 0 , e 0 e j = 0 u = e 0 , e 0 e j = 0 u=e_(0),e_(0)*e_(j)=0\boldsymbol{u}=\boldsymbol{e}_{0}, \boldsymbol{e}_{0} \cdot \boldsymbol{e}_{j}=0u=e0,e0ej=0, and a = 0 a = 0 a=0\boldsymbol{a}=0a=0 for the observer; so Fermi-Walker transport in this case reduces to parallel transport along e 0 e 0 e_(0)\boldsymbol{e}_{0}e0 : thus 0 e j = 0 0 e j = 0 grad_(0)e_(j)=0\boldsymbol{\nabla}_{0} \boldsymbol{e}_{j}=00ej=0. This is, indeed, how e j e j e_(j)\boldsymbol{e}_{j}ej is transported through P o P o P_(o)\mathscr{P}_{o}Po, because
0 e j = Γ α j 0 e α = 0 at P o . 0 e j = Γ α j 0 e α = 0  at  P o . grad_(0)e_(j)=Gamma^(alpha)_(j0)e_(alpha)=0" at "P_(o).\boldsymbol{\nabla}_{0} \boldsymbol{e}_{j}=\Gamma^{\alpha}{ }_{j 0} \boldsymbol{e}_{\alpha}=0 \text { at } \mathscr{P}_{o} .0ej=Γαj0eα=0 at Po.

§8.7. GEODESIC DEVIATION AND THE RIEMANN CURVATURE TENSOR

"Gravitation is a manifestation of spacetime curvature, and that curvature shows up in the deviation of one geodesic from a nearby geodesic (relative acceleration of test particles)." To make this statement precise, first quantify the "deviation" or "relative acceleration" of neighboring geodesics.
Focus attention on a family of geodesics P ( λ , n ) P ( λ , n ) P(lambda,n)\mathscr{P}(\lambda, n)P(λ,n); see Figure 8.4. The smoothly varying parameter n n nnn ("selector parameter") distinguishes one geodesic from the next. For fixed n , P ( λ , n ) n , P ( λ , n ) n,P(lambda,n)n, \mathscr{P}(\lambda, n)n,P(λ,n) is a geodesic with affine parameter λ λ lambda\lambdaλ and with tangent vector
(8.40) u = P / λ ; (8.40) u = P / λ ; {:(8.40)u=delP//del lambda;:}\begin{equation*} \boldsymbol{u}=\partial \mathscr{P} / \partial \lambda ; \tag{8.40} \end{equation*}(8.40)u=P/λ;
thus u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 (geodesic equation). The vector
(8.41) n P / n (8.41) n P / n {:(8.41)n-=delP//del n:}\begin{equation*} \boldsymbol{n} \equiv \partial \mathscr{P} / \partial n \tag{8.41} \end{equation*}(8.41)nP/n
measures the separation between points with the same value of λ λ lambda\lambdaλ on neighboring geodesics.
An observer falling freely along the "fiducial geodesic" n = 0 n = 0 n=0n=0n=0 watches a test particle fall along the "test geodesic" n = 1 n = 1 n=1n=1n=1. The velocity of the test particle relative
Figure 8.4.
A family of geodesics P ( λ , n ) P ( λ , n ) P(lambda,n)\mathscr{P}(\lambda, n)P(λ,n). The selector parameter n n nnn tells "which" geodesic; the affine parameter λ λ lambda\lambdaλ tells "where" on a given geodesic. The separation vector n P / n n P / n n-=delP//del n\boldsymbol{n} \equiv \partial \mathscr{P} / \partial nnP/n at a point P ( λ , 0 ) P ( λ , 0 ) P(lambda,0)\mathscr{\mathscr { P }}(\lambda, 0)P(λ,0) along the fiducial geodesic, n = 0 n = 0 n=0n=0n=0, reaches (approximately) to the point C ( λ , 1 ) C ( λ , 1 ) C(lambda,1)\mathscr{\mathscr { C }}(\lambda, 1)C(λ,1) with the same value of λ λ lambda\lambdaλ on the test geodesic, n = 1 n = 1 n=1n=1n=1.
to him he quantifies by u n u n grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}un. This relative velocity, like the separation vector n n n\boldsymbol{n}n, is an arbitrary "initial condition." Not arbitrary, however, is the "relative acceleration," u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}uun of the test particle relative to the observer (see Boxes 11.2 and 11.3). It would be zero in flat spacetime. In curved spacetime, it is given by
(8.42) u u n + Riemann ( , u , n , u ) = 0 (8.42) u u n + Riemann ( , u , n , u ) = 0 {:(8.42)grad_(u)grad_(u)n+Riemann(dots","u","n","u)=0:}\begin{equation*} \nabla_{u} \nabla_{u} n+\operatorname{Riemann}(\ldots, u, n, u)=0 \tag{8.42} \end{equation*}(8.42)uun+Riemann(,u,n,u)=0
or, in component notation,
(8.43) D 2 n α d λ 2 + R α β γ δ u β n γ u δ = 0 . (8.43) D 2 n α d λ 2 + R α β γ δ u β n γ u δ = 0 . {:(8.43)(D^(2)n^(alpha))/(dlambda^(2))+R^(alpha)_(beta gamma delta)u^(beta)n^(gamma)u^(delta)=0.:}\begin{equation*} \frac{D^{2} n^{\alpha}}{d \lambda^{2}}+R^{\alpha}{ }_{\beta \gamma \delta} u^{\beta} n^{\gamma} u^{\delta}=0 . \tag{8.43} \end{equation*}(8.43)D2nαdλ2+Rαβγδuβnγuδ=0.
This equation serves as a definition of the "Riemann curvature tensor;" and it can also be used to derive the following expressions for the components of Riemann
Riemann curvature tensor defined by relative acceleration of geodesics in a coordinate basis:
(8.44) R β γ δ α = d x α , [ γ , δ ] e β = Γ α β δ x γ Γ α β γ x δ + Γ α μ γ Γ μ β δ Γ α μ δ Γ μ β γ (8.44) R β γ δ α = d x α , γ , δ e β = Γ α β δ x γ Γ α β γ x δ + Γ α μ γ Γ μ β δ Γ α μ δ Γ μ β γ {:[(8.44)R_(beta gamma delta)^(alpha)=(:dx^(alpha),[grad_(gamma),grad_(delta)]e_(beta):)],[=(delGamma^(alpha)_(beta delta))/(delx^(gamma))-(delGamma^(alpha)_(beta gamma))/(delx^(delta))+Gamma^(alpha)_(mu gamma)Gamma^(mu)_(beta delta)-Gamma^(alpha)_(mu delta)Gamma^(mu)_(beta gamma)]:}\begin{align*} R_{\beta \gamma \delta}^{\alpha} & =\left\langle\boldsymbol{d} x^{\alpha},\left[\boldsymbol{\nabla}_{\gamma}, \boldsymbol{\nabla}_{\delta}\right] \boldsymbol{e}_{\beta}\right\rangle \tag{8.44}\\ & =\frac{\partial \Gamma^{\alpha}{ }_{\beta \delta}}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\beta \gamma}}{\partial x^{\delta}}+\Gamma^{\alpha}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta}-\Gamma^{\alpha}{ }_{\mu \delta} \Gamma^{\mu}{ }_{\beta \gamma} \end{align*}(8.44)Rβγδα=dxα,[γ,δ]eβ=ΓαβδxγΓαβγxδ+ΓαμγΓμβδΓαμδΓμβγ
Effects of curvature
Symmetries of Riemann
(For proof, read Box 11.4, Box 11.5, and exercise 11.3, in that order.) For a glimpse of the man who first analyzed the curvature of spaces with three and more dimensions, see Box 8.5.
Spacetime curvature causes not only geodesic deviation, but also route dependence in parallel transport (parallel transport around a closed curve changes a vector or tensor-Box 11.7); it causes covariant derivatives to fail to commute [equation (8.44)]; and it prevents the existence of a global Lorentz coordinate system (§11.5).
At first sight one might think Riemann has 4 × 4 × 4 × 4 = 256 4 × 4 × 4 × 4 = 256 4xx4xx4xx4=2564 \times 4 \times 4 \times 4=2564×4×4×4=256 independent components. But closer examination (§13.5) reveals a variety of symmetries
(8.45) R α β γ δ = R [ α β ] [ γ δ ] = R [ γ δ ] [ α β ] , R [ α β γ δ ] = 0 , R α [ β γ δ ] = 0 (8.45) R α β γ δ = R [ α β ] [ γ δ ] = R [ γ δ ] [ α β ] , R [ α β γ δ ] = 0 , R α [ β γ δ ] = 0 {:(8.45)R_(alpha beta gamma delta)=R_([alpha beta][gamma delta])=R_([gamma delta][alpha beta])","quadR_([alpha beta gamma delta])=0","quadR_(alpha[beta gamma delta])=0:}\begin{equation*} R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta][\gamma \delta]}=R_{[\gamma \delta][\alpha \beta]}, \quad R_{[\alpha \beta \gamma \delta]}=0, \quad R_{\alpha[\beta \gamma \delta]}=0 \tag{8.45} \end{equation*}(8.45)Rαβγδ=R[αβ][γδ]=R[γδ][αβ],R[αβγδ]=0,Rα[βγδ]=0

Box 8.5 GEORG FRIEDRICH BERNHARD RIEMANN

September 17, 1826, Breselenz, Hanover-July 20, 1866, Selasca, Lake Maggiore
With his famous doctoral thesis of 1851, "Foundations for a general theory of functions of a single complex variable," Riemann founded one branch of modern mathematics (the theory of Riemann surfaces); and with his famous lecture of three years later founded another (Riemannian geometry). These and other writings will be found in his collected works, edited by H. Weber (1953).
"The properties which distinguish space from other conceivable triply-extended magnitudes are only to be deduced from experience. ... At every point the three-directional measure of curvature can have an arbitrary value if only the effective curvature of every measurable region of space does not differ noticeably from zero." [G. F. B. Riemann, "On the hypotheses that lie at the foundations of geometry," Habilitationsvorlesung of June 10, 1854, on entry into the philosophical faculty of the University of Göttingen.]
Dying of tuberculosis twelve years later, occu-
(antisymmetry on first two indices; antisymmetry on last two; symmetry under exchange of first pair with last pair; vanishing of completely antisymmetric parts). These reduce Riemann (in four dimensions) from 256 to 20 independent components.
Besides these algebraic symmetries, Riemann possesses differential symmetries called "Bianchi identities,"
(8.46) R β [ λ μ ; v ] α = 0 , (8.46) R β [ λ μ ; v ] α = 0 , {:(8.46)R_(beta[lambda mu;v])^(alpha)=0",":}\begin{equation*} R_{\beta[\lambda \mu ; v]}^{\alpha}=0, \tag{8.46} \end{equation*}(8.46)Rβ[λμ;v]α=0,
Bianchi identities
which have deep geometric significance (Chapter 15).
From Riemann one can form several other curvature tensors by contraction. The easiest to form are the "Ricci curvature tensor,"
pied with an attempt at a unified explanation of gravity and electromagnetism, Riemann communicated to Betti his system of characterization of multiply-connected topologies (which opened the door to the view of electric charge as "lines of force trapped in the topology of space"), making use of numbers that today are named after Betti but that are identified with a symbol, R n R n R_(n)R_{n}Rn, that honors Riemann.
"A more detailed scrutiny of a surface might disclose that what we had considered an elementary piece in reality has tiny handles attached to it which change the connectivity character of the piece, and that a microscope of ever greater magnification would reveal ever new topological complications of this type, ad infinitum. The Riemann point of view allows, also for real space, topological conditions entirely different from those realized by Euclidean space. I believe that only on the basis of the freer and more general conception of geometry which had been brought out by the development of mathematics during the last century, and with an open mind for the imaginative possibilities which it has revealed, can a philosophically fruitful
attack upon the space problem be undertaken." H. Weyl (1949, p. 91).
"But ... physicists were still far removed from such a way of thinking; space was still, for them, a rigid, homogeneous something, susceptible of no change or conditions. Only the genius of Riemann, solitary and uncomprehended, had already won its way by the middle of the last century to a new conception of space, in which space was deprived of its rigidity, and in which its power to take part in physical events was recognized as possible." A. Einstein (1934, p. 68 ).
Riemann formulated the first known model for superspace (for which see Chapter 43), a superspace built, however, not of the totality of all 3geometries with positive definite Riemannian metric (the dynamic arena of Einstein's general relativity), but of all conformally equivalent closed Riemannian 2-geometries of the same topology, a type of superspace known today as Teichmüller space, for more on Riemann's contributions to which and the subsequent development of which, see the chapters by L. Bers and J. A. Wheeler in Gilbert and Newton (1970).
Ricci curvature tensor
Scalar curvature
Einstein curvature tensor
Contracted Bianchi identities
(8.47) R μ ν R α μ α ν = Γ μ ν , α α Γ μ α , ν α + Γ β α α Γ μ ν β Γ β ν α Γ μ α β , (8.47) R μ ν R α μ α ν = Γ μ ν , α α Γ μ α , ν α + Γ β α α Γ μ ν β Γ β ν α Γ μ α β , {:(8.47)R_(mu nu)-=R^(alpha)_(mu alpha nu)=Gamma_(mu nu,alpha)^(alpha)-Gamma_(mu alpha,nu)^(alpha)+Gamma_(beta alpha)^(alpha)Gamma_(mu nu)^(beta)-Gamma_(beta nu)^(alpha)Gamma_(mu alpha)^(beta)",":}\begin{equation*} R_{\mu \nu} \equiv R^{\alpha}{ }_{\mu \alpha \nu}=\Gamma_{\mu \nu, \alpha}^{\alpha}-\Gamma_{\mu \alpha, \nu}^{\alpha}+\Gamma_{\beta \alpha}^{\alpha} \Gamma_{\mu \nu}^{\beta}-\Gamma_{\beta \nu}^{\alpha} \Gamma_{\mu \alpha}^{\beta}, \tag{8.47} \end{equation*}(8.47)RμνRαμαν=Γμν,ααΓμα,να+ΓβααΓμνβΓβναΓμαβ,
and the "scalar curvature,"
(8.48) R R μ μ (8.48) R R μ μ {:(8.48)R-=R^(mu)_(mu):}\begin{equation*} R \equiv R^{\mu}{ }_{\mu} \tag{8.48} \end{equation*}(8.48)RRμμ
But of much greater geometric significance is the "Einstein curvature tensor"
(8.49) G ν μ 1 2 ε μ α β γ R β γ ρ σ 1 2 ε ν α ρ σ = R ν μ 1 2 δ μ ν R . (8.49) G ν μ 1 2 ε μ α β γ R β γ ρ σ 1 2 ε ν α ρ σ = R ν μ 1 2 δ μ ν R . {:(8.49)G_(nu)^(mu)-=(1)/(2)epsi^(mu alpha beta gamma)R_(beta gamma)^(rho sigma)(1)/(2)epsi_(nu alpha rho sigma)=R_(nu)^(mu)-(1)/(2)delta^(mu)_(nu)R.:}\begin{equation*} G_{\nu}^{\mu} \equiv \frac{1}{2} \varepsilon^{\mu \alpha \beta \gamma} R_{\beta \gamma}{ }^{\rho \sigma} \frac{1}{2} \varepsilon_{\nu \alpha \rho \sigma}=R_{\nu}^{\mu}-\frac{1}{2} \delta^{\mu}{ }_{\nu} R . \tag{8.49} \end{equation*}(8.49)Gνμ12εμαβγRβγρσ12εναρσ=Rνμ12δμνR.
Of all second-rank curvature tensors one can form by contracting Riemann, only
Einstein = G = G =G=\boldsymbol{G}=G retains part of the Bianchi identities (8.46): it satisfies
(8.50) G ; ν μ ν = 0 (8.50) G ; ν μ ν = 0 {:(8.50)G_(;nu)^(mu nu)=0:}\begin{equation*} G_{; \nu}^{\mu \nu}=0 \tag{8.50} \end{equation*}(8.50)G;νμν=0
For the beautiful geometric meaning of these "contracted Bianchi identities" ("the boundary of a boundary is zero"), see Chapter 15 .
Box 8.6 summarizes the above equations describing curvature, as well as the fundamental equations for covariant derivatives.

EXERCISE

[The following exercises from Track 2 are appropriate for the Track-1 reader who wishes to solidfy his understanding of curvature: 11.6 , 11.9 , 11.10 , 13.7 11 11.6 , 11.9 , 11.10 , 13.7 11 11.6,11.9,11.10,13.7-1111.6,11.9,11.10,13.7-1111.6,11.9,11.10,13.711, and 14.3.]

Exercise 8.16. SOME USEFUL FORMULAS IN COORDINATE FRAMES

In any coordinate frame, define g g ggg to be the determinant of the matrix g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ [equation 8.11]. Derive the following relations, valid in any coordinate frame.
(a) Contraction of connection coefficients:
(8.51a) Γ β α α = ( ln g ) , β (8.51a) Γ β α α = ( ln g ) , β {:(8.51a)Gamma_(beta alpha)^(alpha)=(ln sqrt(-g))_(,beta):}\begin{equation*} \Gamma_{\beta \alpha}^{\alpha}=(\ln \sqrt{-g})_{, \beta} \tag{8.51a} \end{equation*}(8.51a)Γβαα=(lng),β
[Hint: Use the results of exercise 5.5.]
(b) Components of Ricci tensor:
(8.51b) R α β = 1 g ( g Γ α β μ ) , μ ( ln g ) , α β Γ μ ν α Γ v β μ . (8.51b) R α β = 1 g g Γ α β μ , μ ( ln g ) , α β Γ μ ν α Γ v β μ . {:(8.51b)R_(alpha beta)=(1)/(sqrt(-g))(sqrt(-g)Gamma_(alpha beta)^(mu))_(,mu)-(ln sqrt(-g))_(,alpha beta)-Gamma^(mu)_(nu alpha)Gamma^(v)_(beta mu).:}\begin{equation*} R_{\alpha \beta}=\frac{1}{\sqrt{-g}}\left(\sqrt{-g} \Gamma_{\alpha \beta}^{\mu}\right)_{, \mu}-(\ln \sqrt{-g})_{, \alpha \beta}-\Gamma^{\mu}{ }_{\nu \alpha} \Gamma^{v}{ }_{\beta \mu} . \tag{8.51b} \end{equation*}(8.51b)Rαβ=1g(gΓαβμ),μ(lng),αβΓμναΓvβμ.
(c) Divergence of a vector A α A α A^(alpha)A^{\alpha}Aα or antisymmetric tensor F α β F α β F^(alpha beta)F^{\alpha \beta}Fαβ :
(8.51c) A ; α α = 1 g ( g A α ) , α , F ; β α β = 1 g ( g F α β ) , β . (8.51c) A ; α α = 1 g g A α , α , F ; β α β = 1 g g F α β , β . {:(8.51c)A_(;alpha)^(alpha)=(1)/(sqrt(-g))(sqrt(-g)A^(alpha))_(,alpha)","quadF_(;beta)^(alpha beta)=(1)/(sqrt(-g))(sqrt(-g)F^(alpha beta))_(,beta).:}\begin{equation*} A_{; \alpha}^{\alpha}=\frac{1}{\sqrt{-g}}\left(\sqrt{-g} A^{\alpha}\right)_{, \alpha}, \quad F_{; \beta}^{\alpha \beta}=\frac{1}{\sqrt{-g}}\left(\sqrt{-g} F^{\alpha \beta}\right)_{, \beta} . \tag{8.51c} \end{equation*}(8.51c)A;αα=1g(gAα),α,F;βαβ=1g(gFαβ),β.
(d) Integral of a scalar field Ψ Ψ Psi\PsiΨ over the proper volume of a 4-dimensional region V V V\mathscr{V}V :
(8.51~d) γ Ψ d (proper volume ) = γ Ψ g d t d x d y d z (8.51~d) γ Ψ d  (proper volume  = γ Ψ g d t d x d y d z {:(8.51~d){:int_(gamma)Psi d" (proper volume ")=int_(gamma)Psisqrt(-g)dtdxdydz:}\begin{equation*} \left.\int_{\gamma} \Psi d \text { (proper volume }\right)=\int_{\gamma} \Psi \sqrt{-g} d t d x d y d z \tag{8.51~d} \end{equation*}(8.51~d)γΨd (proper volume )=γΨgdtdxdydz
[Hint: In a local Lorentz frame, d d ddd (proper volume) = d t ^ d x ^ d y ^ d z ^ = d t ^ d x ^ d y ^ d z ^ =d hat(t)d hat(x)d hat(y)d hat(z)=d \hat{t} d \hat{x} d \hat{y} d \hat{z}=dt^dx^dy^dz^. Use a Jacobian to transform this volume element to the given coordinate frame, and prove from the transformation law
g α β = x μ ^ x α x v ^ x β η μ v g α β = x μ ^ x α x v ^ x β η μ v g_(alpha beta)=(delx^( hat(mu)))/(delx^(alpha))(delx^( hat(v)))/(delx^(beta))eta_(mu v)g_{\alpha \beta}=\frac{\partial x^{\hat{\mu}}}{\partial x^{\alpha}} \frac{\partial x^{\hat{v}}}{\partial x^{\beta}} \eta_{\mu v}gαβ=xμ^xαxv^xβημv
that the Jacobian is equal to g g sqrt(-g)\sqrt{-g}g.]
Box 8.6 COVARIANT DERIVATIVE AND CURVATURE: FUNDAMENTAL EQUATIONS

Box 8.6 (continued)

Entity Abstract notation Component noiation
Geodesic Equation u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 d 2 x α / d λ 2 + Γ α μ v ( d x μ / d λ ) ( d x ν / d λ ) = 0 in a coordinate basis d 2 x α / d λ 2 + Γ α μ v d x μ / d λ d x ν / d λ = 0  in a coordinate basis  {:[d^(2)x^(alpha)//dlambda^(2)+Gamma^(alpha)_(mu v)(dx^(mu)//d lambda)(dx^(nu)//d lambda)=0],[quad" in a coordinate basis "]:}\begin{aligned} & d^{2} x^{\alpha} / d \lambda^{2}+\Gamma^{\alpha}{ }_{\mu v}\left(d x^{\mu} / d \lambda\right)\left(d x^{\nu} / d \lambda\right)=0 \\ & \quad \text { in a coordinate basis } \end{aligned}d2xα/dλ2+Γαμv(dxμ/dλ)(dxν/dλ)=0 in a coordinate basis 
Riemann Curvature Tensor Riemann ( σ , C , A , B ) σ , R ( A , B ) C ( σ , C , A , B ) σ , R ( A , B ) C (sigma,C,A,B)-=(:sigma,R(A,B)C:)(\sigma, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B}) \equiv\langle\sigma, \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}\rangle(σ,C,A,B)σ,R(A,B)C R ( A , B ) [ A , B ] [ A , B ] R ( A , B ) A , B [ A , B ] R(A,B)-=[grad_(A),grad_(B)]-grad_([A,B])\mathscr{R}(A, B) \equiv\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{A}, \boldsymbol{B}]}R(A,B)[A,B][A,B] (not track-one formulas; see Chapter 11)
R α β γ δ = Γ α β δ x γ Γ α β γ x δ + Γ α μ γ Γ μ β δ Γ α μ δ Γ μ β γ R α β γ δ = Γ α β δ x γ Γ α β γ x δ + Γ α μ γ Γ μ β δ Γ α μ δ Γ μ β γ {:[R^(alpha)_(beta gamma delta)=(delGamma^(alpha)_(beta delta))/(delx^(gamma))-(delGamma^(alpha)_(beta gamma))/(delx^(delta))],[+Gamma^(alpha)_(mu gamma)Gamma^(mu)_(beta delta)-Gamma^(alpha)_(mu delta)Gamma^(mu)_(beta gamma)]:}\begin{aligned} R^{\alpha}{ }_{\beta \gamma \delta}= & \frac{\partial \Gamma^{\alpha}{ }_{\beta \delta}}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\beta \gamma}}{\partial x^{\delta}} \\ & +\Gamma^{\alpha}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta}-\Gamma^{\alpha}{ }_{\mu \delta} \Gamma^{\mu}{ }_{\beta \gamma} \end{aligned}Rαβγδ=ΓαβδxγΓαβγxδ+ΓαμγΓμβδΓαμδΓμβγ
in coordinate frame
[see equation (11.13) for formula in noncoordinate frame]
"R^(alpha)_(beta gamma delta)=(delGamma^(alpha)_(beta delta))/(delx^(gamma))-(delGamma^(alpha)_(beta gamma))/(delx^(delta)) +Gamma^(alpha)_(mu gamma)Gamma^(mu)_(beta delta)-Gamma^(alpha)_(mu delta)Gamma^(mu)_(beta gamma)" in coordinate frame [see equation (11.13) for formula in noncoordinate frame]| $\begin{aligned} R^{\alpha}{ }_{\beta \gamma \delta}= & \frac{\partial \Gamma^{\alpha}{ }_{\beta \delta}}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\beta \gamma}}{\partial x^{\delta}} \\ & +\Gamma^{\alpha}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta}-\Gamma^{\alpha}{ }_{\mu \delta} \Gamma^{\mu}{ }_{\beta \gamma} \end{aligned}$ | | :--- | | in coordinate frame | | [see equation (11.13) for formula in noncoordinate frame] |
Ricci Curvature
Tensor
Ricci Curvature Tensor| Ricci Curvature | | :--- | | Tensor |
Ricci = contraction on slots 1 and 3 of Riemann  Ricci  =  contraction on slots  1  and  3  of Riemann  {:[" Ricci "=" contraction on slots "],[1" and "3" of Riemann "]:}\begin{aligned} \text { Ricci }= & \text { contraction on slots } \\ & 1 \text { and } 3 \text { of Riemann } \end{aligned} Ricci = contraction on slots 1 and 3 of Riemann 
R μ ν = R α μ a v = Γ α μ v , α Γ α μ α , v + Γ α β α Γ β μ r Γ α β ν Γ β μ α R μ ν = R α μ a v = Γ α μ v , α Γ α μ α , v + Γ α β α Γ β μ r Γ α β ν Γ β μ α {:[R_(mu nu)=R^(alpha)_(mu av)=Gamma^(alpha)_(mu v,alpha)-Gamma^(alpha)_(mu alpha,v)+Gamma^(alpha)_(beta alpha)Gamma^(beta)_(mu r)],[-Gamma^(alpha)_(beta nu)Gamma^(beta)_(mu alpha)]:}\begin{aligned} R_{\mu \nu}=R^{\alpha}{ }_{\mu a v}=\Gamma^{\alpha}{ }_{\mu v, \alpha}-\Gamma^{\alpha}{ }_{\mu \alpha, v} & +\Gamma^{\alpha}{ }_{\beta \alpha} \Gamma^{\beta}{ }_{\mu r} \\ & -\Gamma^{\alpha}{ }_{\beta \nu} \Gamma^{\beta}{ }_{\mu \alpha} \end{aligned}Rμν=Rαμav=Γαμv,αΓαμα,v+ΓαβαΓβμrΓαβνΓβμα
in coordinate frame
"R_(mu nu)=R^(alpha)_(mu av)=Gamma^(alpha)_(mu v,alpha)-Gamma^(alpha)_(mu alpha,v)+Gamma^(alpha)_(beta alpha)Gamma^(beta)_(mu r) -Gamma^(alpha)_(beta nu)Gamma^(beta)_(mu alpha)" in coordinate frame| $\begin{aligned} R_{\mu \nu}=R^{\alpha}{ }_{\mu a v}=\Gamma^{\alpha}{ }_{\mu v, \alpha}-\Gamma^{\alpha}{ }_{\mu \alpha, v} & +\Gamma^{\alpha}{ }_{\beta \alpha} \Gamma^{\beta}{ }_{\mu r} \\ & -\Gamma^{\alpha}{ }_{\beta \nu} \Gamma^{\beta}{ }_{\mu \alpha} \end{aligned}$ | | :--- | | in coordinate frame |
*Curvature Scalar R = ( R = ( R=(R=(R=( contraction of Ricci) R = R α α R = R α α R=R^(alpha)_(alpha)R=R^{\alpha}{ }_{\alpha}R=Rαα
*Einstein Curvature Tensor G = R i c c i 1 2 g R G = R i c c i 1 2 g R G=Ricci-(1)/(2)gR\boldsymbol{G}=\boldsymbol{R i c c i}-\frac{1}{2} \boldsymbol{g} RG=Ricci12gR
G α β = R α β 1 2 g α β R G α β = R α β 1 2 g α β R G_(alpha beta)=R_(alpha beta)-(1)/(2)g_(alpha beta)RG_{\alpha \beta}=R_{\alpha \beta}-\frac{1}{2} g_{\alpha \beta} RGαβ=Rαβ12gαβR
Useful formulas for computing G α β G α β G^(alpha)_(beta)G^{\alpha}{ }_{\beta}Gαβ (derived in §14.2): G 0 0 = ( R 12 12 + R 23 23 + R 31 31 ) , G 0 1 = R 02 12 + R 03 13 , G 1 1 = ( R 02 02 + R 03 03 + R 23 23 ) , G 1 2 = R 10 20 + R 13 23 , etc. G 0 0 = R 12 12 + R 23 23 + R 31 31 , G 0 1 = R 02 12 + R 03 13 , G 1 1 = R 02 02 + R 03 03 + R 23 23 , G 1 2 = R 10 20 + R 13 23 ,  etc.  {:[G^(0)_(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31))","],[G^(0)_(1)=R^(02)_(12)+R^(03)_(13)","],[G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23))","],[G^(1)_(2)=R^(10)_(20)+R^(13)_(23)","" etc. "]:}\begin{aligned} & G^{0}{ }_{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right), \\ & G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}, \\ & G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23}\right), \\ & G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}, \text { etc. } \end{aligned}G00=(R1212+R2323+R3131),G01=R0212+R0313,G11=(R0202+R0303+R2323),G12=R1020+R1323, etc. 
G_(alpha beta)=R_(alpha beta)-(1)/(2)g_(alpha beta)R Useful formulas for computing G^(alpha)_(beta) (derived in §14.2): "G^(0)_(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31)), G^(0)_(1)=R^(02)_(12)+R^(03)_(13), G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23)), G^(1)_(2)=R^(10)_(20)+R^(13)_(23), etc. "| $G_{\alpha \beta}=R_{\alpha \beta}-\frac{1}{2} g_{\alpha \beta} R$ | | :--- | | Useful formulas for computing $G^{\alpha}{ }_{\beta}$ (derived in §14.2): $\begin{aligned} & G^{0}{ }_{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right), \\ & G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}, \\ & G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23}\right), \\ & G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}, \text { etc. } \end{aligned}$ |
*Symmetries of Curvature Tensors R α β γ δ = R [ α β ] γ δ ] = R [ γ δ [ α β ] , R [ α β γ δ ] = 0 , R α [ β γ δ ] = 0 R α β = R ( α β ) , G α β = G ( α β ) R α β γ δ = R [ α β ] γ δ ] = R [ γ δ [ α β ] , R [ α β γ δ ] = 0 , R α [ β γ δ ] = 0 R α β = R ( α β ) , G α β = G ( α β ) {:[R_(alpha beta gamma delta)=R_([alpha beta]∣gamma delta])=R_([gamma delta∣[alpha beta])","R_([alpha beta gamma delta])=0","R_(alpha[beta gamma delta])=0],[R_(alpha beta)=R_((alpha beta))","G_(alpha beta)=G_((alpha beta))]:}\begin{aligned} & R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta] \mid \gamma \delta]}=R_{[\gamma \delta \mid[\alpha \beta]}, R_{[\alpha \beta \gamma \delta]}=0, R_{\alpha[\beta \gamma \delta]}=0 \\ & R_{\alpha \beta}=R_{(\alpha \beta)}, G_{\alpha \beta}=G_{(\alpha \beta)} \end{aligned}Rαβγδ=R[αβ]γδ]=R[γδ[αβ],R[αβγδ]=0,Rα[βγδ]=0Rαβ=R(αβ),Gαβ=G(αβ)
Bianchi Identities R α β [ μ ν ; N ] = 0 R α β [ μ ν ; N ] = 0 R^(alpha)_(beta[mu nu;N])=0R^{\alpha}{ }_{\beta[\mu \nu ; \mathrm{N}]}=0Rαβ[μν;N]=0
* Contracted
Bianchi
Identities
* Contracted Bianchi Identities| * Contracted | | :--- | | Bianchi | | Identities |
G α β ; β = 0 G α β ; β = 0 G^(alpha beta)_(;beta)=0G^{\alpha \beta}{ }_{; \beta}=0Gαβ;β=0
Geodesic Deviation u u n + Riemann ( , u , n , u ) = 0 u u n + Riemann ( , u , n , u ) = 0 grad_(u)grad_(u)n+Riemann(dots,u,n,u)=0\nabla_{u} \nabla_{u} \boldsymbol{n}+\operatorname{Riemann}(\ldots, u, n, u)=0uun+Riemann(,u,n,u)=0 D 2 n α d λ 2 + R α β γ δ u β n γ u δ = 0 D 2 n α d λ 2 + R α β γ δ u β n γ u δ = 0 (D^(2)n^(alpha))/(dlambda^(2))+R^(alpha)_(beta gamma delta)u^(beta)n^(gamma)u^(delta)=0\frac{D^{2} n^{\alpha}}{d \lambda^{2}}+R^{\alpha}{ }_{\beta \gamma \delta} u^{\beta} n^{\gamma} u^{\delta}=0D2nαdλ2+Rαβγδuβnγuδ=0
Parallel Transport around closed curve (811.4)
δ A + R i e m a n n ( A , u , v ) = 0 δ A + R i e m a n n ( A , u , v ) = 0 delta A+Riemann(dots A,u,v)=0\delta \boldsymbol{A}+\boldsymbol{R i e m a n n}(\ldots \boldsymbol{A}, \boldsymbol{u}, \boldsymbol{v})=0δA+Riemann(A,u,v)=0
if u , v u , v u,v\boldsymbol{u}, \boldsymbol{v}u,v are edges of curve
delta A+Riemann(dots A,u,v)=0 if u,v are edges of curve| $\delta \boldsymbol{A}+\boldsymbol{R i e m a n n}(\ldots \boldsymbol{A}, \boldsymbol{u}, \boldsymbol{v})=0$ | | :--- | | if $\boldsymbol{u}, \boldsymbol{v}$ are edges of curve |
δ A α + R α β γ δ A β u γ v δ = 0 δ A α + R α β γ δ A β u γ v δ = 0 deltaA^(alpha)+R^(alpha)_(beta gamma delta)A^(beta)u^(gamma)v^(delta)=0\delta A^{\alpha}+R^{\alpha}{ }_{\beta \gamma \delta} A^{\beta} u^{\gamma} v^{\delta}=0δAα+RαβγδAβuγvδ=0
Entity Abstract notation Component noiation Geodesic Equation grad_(u)u=0 "d^(2)x^(alpha)//dlambda^(2)+Gamma^(alpha)_(mu v)(dx^(mu)//d lambda)(dx^(nu)//d lambda)=0 quad in a coordinate basis " Riemann Curvature Tensor Riemann (sigma,C,A,B)-=(:sigma,R(A,B)C:) R(A,B)-=[grad_(A),grad_(B)]-grad_([A,B]) (not track-one formulas; see Chapter 11) ""R^(alpha)_(beta gamma delta)=(delGamma^(alpha)_(beta delta))/(delx^(gamma))-(delGamma^(alpha)_(beta gamma))/(delx^(delta)) +Gamma^(alpha)_(mu gamma)Gamma^(mu)_(beta delta)-Gamma^(alpha)_(mu delta)Gamma^(mu)_(beta gamma)" in coordinate frame [see equation (11.13) for formula in noncoordinate frame]" "Ricci Curvature Tensor" " Ricci = contraction on slots 1 and 3 of Riemann " ""R_(mu nu)=R^(alpha)_(mu av)=Gamma^(alpha)_(mu v,alpha)-Gamma^(alpha)_(mu alpha,v)+Gamma^(alpha)_(beta alpha)Gamma^(beta)_(mu r) -Gamma^(alpha)_(beta nu)Gamma^(beta)_(mu alpha)" in coordinate frame" *Curvature Scalar R=( contraction of Ricci) R=R^(alpha)_(alpha) *Einstein Curvature Tensor G=Ricci-(1)/(2)gR "G_(alpha beta)=R_(alpha beta)-(1)/(2)g_(alpha beta)R Useful formulas for computing G^(alpha)_(beta) (derived in §14.2): "G^(0)_(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31)), G^(0)_(1)=R^(02)_(12)+R^(03)_(13), G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23)), G^(1)_(2)=R^(10)_(20)+R^(13)_(23), etc. "" *Symmetries of Curvature Tensors "R_(alpha beta gamma delta)=R_([alpha beta]∣gamma delta])=R_([gamma delta∣[alpha beta]),R_([alpha beta gamma delta])=0,R_(alpha[beta gamma delta])=0 R_(alpha beta)=R_((alpha beta)),G_(alpha beta)=G_((alpha beta))" Bianchi Identities R^(alpha)_(beta[mu nu;N])=0 "* Contracted Bianchi Identities" G^(alpha beta)_(;beta)=0 Geodesic Deviation grad_(u)grad_(u)n+Riemann(dots,u,n,u)=0 (D^(2)n^(alpha))/(dlambda^(2))+R^(alpha)_(beta gamma delta)u^(beta)n^(gamma)u^(delta)=0 Parallel Transport around closed curve (811.4) "delta A+Riemann(dots A,u,v)=0 if u,v are edges of curve" deltaA^(alpha)+R^(alpha)_(beta gamma delta)A^(beta)u^(gamma)v^(delta)=0| Entity Abstract notation | | Component noiation | | :---: | :---: | :---: | | Geodesic Equation | $\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0$ | $\begin{aligned} & d^{2} x^{\alpha} / d \lambda^{2}+\Gamma^{\alpha}{ }_{\mu v}\left(d x^{\mu} / d \lambda\right)\left(d x^{\nu} / d \lambda\right)=0 \\ & \quad \text { in a coordinate basis } \end{aligned}$ | | Riemann Curvature Tensor | Riemann $(\sigma, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B}) \equiv\langle\sigma, \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}\rangle$ $\mathscr{R}(A, B) \equiv\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{A}, \boldsymbol{B}]}$ (not track-one formulas; see Chapter 11) | $\begin{aligned} R^{\alpha}{ }_{\beta \gamma \delta}= & \frac{\partial \Gamma^{\alpha}{ }_{\beta \delta}}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\beta \gamma}}{\partial x^{\delta}} \\ & +\Gamma^{\alpha}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta}-\Gamma^{\alpha}{ }_{\mu \delta} \Gamma^{\mu}{ }_{\beta \gamma} \end{aligned}$ <br> in coordinate frame <br> [see equation (11.13) for formula in noncoordinate frame] | | Ricci Curvature <br> Tensor | $\begin{aligned} \text { Ricci }= & \text { contraction on slots } \\ & 1 \text { and } 3 \text { of Riemann } \end{aligned}$ | $\begin{aligned} R_{\mu \nu}=R^{\alpha}{ }_{\mu a v}=\Gamma^{\alpha}{ }_{\mu v, \alpha}-\Gamma^{\alpha}{ }_{\mu \alpha, v} & +\Gamma^{\alpha}{ }_{\beta \alpha} \Gamma^{\beta}{ }_{\mu r} \\ & -\Gamma^{\alpha}{ }_{\beta \nu} \Gamma^{\beta}{ }_{\mu \alpha} \end{aligned}$ <br> in coordinate frame | | *Curvature Scalar | $R=($ contraction of Ricci) | $R=R^{\alpha}{ }_{\alpha}$ | | *Einstein Curvature Tensor | $\boldsymbol{G}=\boldsymbol{R i c c i}-\frac{1}{2} \boldsymbol{g} R$ | $G_{\alpha \beta}=R_{\alpha \beta}-\frac{1}{2} g_{\alpha \beta} R$ <br> Useful formulas for computing $G^{\alpha}{ }_{\beta}$ (derived in §14.2): $\begin{aligned} & G^{0}{ }_{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right), \\ & G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}, \\ & G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23}\right), \\ & G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}, \text { etc. } \end{aligned}$ | | *Symmetries of Curvature Tensors | | $\begin{aligned} & R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta] \mid \gamma \delta]}=R_{[\gamma \delta \mid[\alpha \beta]}, R_{[\alpha \beta \gamma \delta]}=0, R_{\alpha[\beta \gamma \delta]}=0 \\ & R_{\alpha \beta}=R_{(\alpha \beta)}, G_{\alpha \beta}=G_{(\alpha \beta)} \end{aligned}$ | | Bianchi Identities | | $R^{\alpha}{ }_{\beta[\mu \nu ; \mathrm{N}]}=0$ | | * Contracted <br> Bianchi <br> Identities | | $G^{\alpha \beta}{ }_{; \beta}=0$ | | Geodesic Deviation | $\nabla_{u} \nabla_{u} \boldsymbol{n}+\operatorname{Riemann}(\ldots, u, n, u)=0$ | $\frac{D^{2} n^{\alpha}}{d \lambda^{2}}+R^{\alpha}{ }_{\beta \gamma \delta} u^{\beta} n^{\gamma} u^{\delta}=0$ | | Parallel Transport around closed curve (811.4) | $\delta \boldsymbol{A}+\boldsymbol{R i e m a n n}(\ldots \boldsymbol{A}, \boldsymbol{u}, \boldsymbol{v})=0$ <br> if $\boldsymbol{u}, \boldsymbol{v}$ are edges of curve | $\delta A^{\alpha}+R^{\alpha}{ }_{\beta \gamma \delta} A^{\beta} u^{\gamma} v^{\delta}=0$ |

CHAPTER

DIFFERENTIAL TOPOLOGY

In analytic geometry, many relations which are independent of any frame must be expressed with respect to some particular frame. It is therefore preferable to devise new methodsmethods which lead directly to intrinsic properties without any mention of coordinates. The development of the topology of general spaces and of the objects which occur in them, as well as the development of the geometry of general metric spaces, are steps in this direction.
KARL MENGER, in Schilpp (1949), p. 467

§9.1. GEOMETRIC OBJECTS IN METRIC-FREE, GEODESIC-FREE SPACETIME

Curved spacetime without metric or geodesics or parallel transport, i.e., "differential topology," is the subject of this easy chapter. It is easy because all the necessary geometric objects (event, curve, vector, 1 -form, tensor) are already familiar from flat spacetime. Yet it is also necessary, because one's viewpoint must be refined when one abandons the Lorentz metric of flat spacetime.

Events

The primitive concept of an event P P P\mathscr{P}P (Figure 1.2) needs no refinement. The essential property here is identifiability, which is not dependent on the Lorentz metric structure of spacetime.
This chapter is entirely Track 2. It depends on no preceding Track-2 material.
It is needed as preparation for
(1) Chapters 10-13 (differential geometry; Newtonian gravity), and
(2) Box 30.1 (mixmaster cosmology).
It will be helpful in
(1) Chapter 14 (calculation of curvature) and in
(2) Chapter 15 (Bianchi identities).
Geometric concepts must be refined
Old definitions of vector break down when metric is abandoned

Curves

Again no refinement. A "curve" P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) is also too primitive to care whether spacetime has a metric-except that, with metric gone, there is no concept of "proper length" along the curve. This is in accord with Newton's theory of gravity, where one talks of the lengths of curves in "space," but never in "spacetime."

Vectors

Here refinement is needed. In special relativity one could dress primitive ("identifiable") events in enough algebraic plumage to talk of vectors as differences P Q P Q P-Q\mathscr{P}-\mathscr{Q}PQ between "algebraic" events. Now the plumage is gone, and the old bilocal ("point for head and point for tail") version of a vector must be replaced by a purely local version ( $ 9.2 $ 9.2 $9.2\$ 9.2$9.2 ). Also vectors cannot be moved around; each vector must be attached to a specific event ( § § 9.2 § § 9.2 §§9.2\S \S 9.2§§9.2 and 9.3).

1-Forms

Almost no refinement needed, except that, with metric gone, there is no way to tell which 1 -form corresponds to a given vector (no way to raise and lower indices), and each 1 -form must be attached to a specific event ( § 9.4 § 9.4 §9.4\S 9.4§9.4 ).

Tensors

Again almost no refinement, except that each slot of a tensor is specific: if it accepts vectors, then it cannot accommodate 1 -forms, and conversely (no raising and lowering of indices); also, each tensor must be attached to a specific event ( $ 9.5 $ 9.5 $9.5\$ 9.5$9.5 ).

§9.2. "VECTOR" AND "DIRECTIONAL DERIVATIVE" REFINED INTO TANGENT VECTOR

Flat spacetime can accommodate several equivalent definitions of a vector (§2.3): a vector is an arrow reaching from an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 to an event Q 0 Q 0 Q_(0)\mathscr{Q}_{0}Q0; it is the parameterized straight line, P ( λ ) = P 0 + λ ( Q 0 P 0 ) P ( λ ) = P 0 + λ Q 0 P 0 P(lambda)=P_(0)+lambda(Q_(0)-P_(0))\mathscr{P}(\lambda)=\mathscr{P}_{0}+\lambda\left(\mathscr{Q}_{0}-\mathscr{P}_{0}\right)P(λ)=P0+λ(Q0P0) extending from P 0 P 0 P_(0)\mathscr{P}_{0}P0 at λ = 0 λ = 0 lambda=0\lambda=0λ=0 to Q 0 Q 0 Q_(0)\mathscr{Q}_{0}Q0 at λ = 1 λ = 1 lambda=1\lambda=1λ=1; it is the rate of change of the point P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) with increasing λ , d P / d λ λ , d P / d λ lambda,dP//d lambda\lambda, d \mathscr{P} / d \lambdaλ,dP/dλ.
With Lorentz metric gone, the "arrow" definition and the "parametrized-straight line" definition must break down. By what route is the arrow or line to be laid out between P 0 P 0 P_(0)\mathscr{P}_{0}P0 and Q 0 Q 0 Q_(0)\mathscr{Q}_{0}Q0 ? There is no concept of straightness; all routes are equally straight or bent.
Such fuzziness forces one to focus on the "rate-of-change-of-point-along-curve"

Box 9.1 TANGENT VECTORS AND TANGENT SPACE

A tangent vector d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ is defined to be "the limit, when N N N longrightarrow ooN \longrightarrow \inftyN, of N N NNN times the displacement of P P P\mathscr{P}P as λ λ lambda\lambdaλ ranges from 0 to 1 / N 1 / N 1//N1 / N1/N." One cannot think of this final displacement d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ as lying in spacetime; fuzziness forbids (no concept of straightness). Instead, one visualizes d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ as lying in a "tangent plane" or "tangent space," which makes contact with spacetime only at P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0), the event where d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ is evaluated. All other tangent vectors at P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0)-e.g., d P / d ρ , d P / d η , d P / d ζ d P / d ρ , d P / d η , d P / d ζ dP//d rho,dP//d eta,dP//d zetad \mathscr{P} / d \rho, d \mathscr{P} / d \eta, d \mathscr{P} / d \zetadP/dρ,dP/dη,dP/dζ-lie in this same tangent space.
To make precise these concepts of tangent space and tangent vector, one may regard spacetime as embedded in a flat space of more than four di-
mensions. One can then perform the limiting process that leads to d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ, using straight arrows in the flat embedding space. The result is a higherdimensional analog of the figure shown above.
But such a treatment is dangerous. It suggests, falsely, that the tangent vector d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ and the tangent space at P 0 P 0 P_(0)\mathscr{P}_{0}P0 depend on how the embedding is done, or depend for their existence on the embedding process. They do not. And to make clear that they do not is one motivation for defining the directional derivative operator " d / d λ d / d λ d//d lambdad / d \lambdad/dλ " to be the tangent vector, rather than using Cartan's more pictorial concept " d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ ".
definition, d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ. It, under the new name "tangent vector," is explored briefly in Box 9.1, and in greater depth in the following paragraphs.
Even " d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ " is a fuzzy definition of tangent vector, most mathematicians would argue. More acceptable, they suggest, is this definition: the tangent vector u u u\boldsymbol{u}u to a a aaa curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) is the directional derivative operator along that curve
(9.1) u = u = ( d / d λ ) along curve (9.1) u = u = ( d / d λ ) along curve  {:(9.1)u=del_(u)=(d//d lambda)_("along curve "):}\begin{equation*} \boldsymbol{u}=\partial_{\boldsymbol{u}}=(d / d \lambda)_{\text {along curve }} \tag{9.1} \end{equation*}(9.1)u=u=(d/dλ)along curve 
Tangent vector equals directional derivative operator? Preposterous! A vector started out as a happy, irresponsible trip from P 0 P 0 P_(0)\mathscr{P}_{0}P0 to Q 0 Q 0 Q_(0)\mathscr{Q}_{0}Q0. It ended up loaded with the social responsibility to tell how something else changes at P 0 P 0 P_(0)\mathscr{P}_{0}P0. At what point did the vector get saddled with this unexpected load? And did it really change its character all that much, as it seems to have done? For an answer, go back and try
Best new definition: "tangent vector equals directional derivative operator"
u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ
to redo the "rate-of-change-of-point" definition, d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ, in the form of a limiting process:
0 . Choose a curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) whose tangent vector d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ at λ = 0 λ = 0 lambda=0\lambda=0λ=0 is desired.
  1. Take the displacement of P P P\mathscr{P}P as λ λ lambda\lambdaλ ranges from 0 to 1 ; that is not d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ.
  2. Take twice the displacement of P P P\mathscr{P}P as λ λ lambda\lambdaλ ranges from 0 to 1 2 1 2 (1)/(2)\frac{1}{2}12; that is not d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ.
    N N NNN. Take N N NNN times the displacement of P P P\mathscr{P}P as λ λ lambda\lambdaλ ranges from 0 to 1 / N 1 / N 1//N1 / N1/N; that is not d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ.
    oo\infty. Take the limit of such displacements as N N N longrightarrow ooN \longrightarrow \inftyN; that is d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ.
    This definition has the virtue that d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ describes the properties of the curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ), not over the huge range from λ = 0 λ = 0 lambda=0\lambda=0λ=0 to λ = 1 λ = 1 lambda=1\lambda=1λ=1, where the curve might be doing wild things, but only in an infinitesimal neighborhood of the point P 0 = P ( 0 ) P 0 = P ( 0 ) P_(0)=P(0)\mathscr{P}_{0}=\mathscr{P}(0)P0=P(0).
The deficiency in this definition is that no meaning is assigned to steps 1 , 2 , 1 , 2 , 1,2,dots1,2, \ldots1,2,, N , N , N,dotsN, \ldotsN,, so there is nothing, yet, to take the limit of. To make each "displacement of P P P\mathscr{P}P " a definite mathematical object in a space where "limit" has a meaning, one can imagine the original manifold to be a low-dimensional surface in some much higher-dimensional flat space. Then P ( 1 / N ) P ( 0 ) P ( 1 / N ) P ( 0 ) P(1//N)-P(0)\mathscr{P}(1 / N)-\mathscr{P}(0)P(1/N)P(0) is just a straight arrow connecting two points, i.e. a segment of a straight line, which, in general, will not lie in the surface itself-see Box 9.1. The resulting mental picture of a tangent vector makes its essential properties beautifully clear, but at the cost of some artifacts. The picture relies on a specific but arbitrary way of embedding the manifold of interest (metricfree spacetime) in an extraneous flat space. In using this picture, one must ignore everything that depends on the peculiarities of the embedding. One must think like the chemist, who uses tinkertoy molecular models to visualize many essential properties of a molecule clearly, but easily ignores artifacts of the model (colors of the atoms, diameters of the pegs, its tendency to collapse) that do not mimic quantummechanical reality.
Élie Cartan's approach to differential geometry, including the d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ idea of a tangent vector, suggests that he always thought of manifolds as embedded in flat spaces this way, and relied on insights that he did not always formalize to separate the essential geometry of these pictures from their embedding-dependent details. Acceptance of his methods of calculation came late. Mathematicians, who mistrusted their own ability to distinguish fact from artifact, exacted this price for acceptance: stop talking about the movement of the point itself, and start dealing only with concrete measurable changes that take place within the manifold, changes in any or all scalar functions f f fff as the point moves. The limiting process then reads:
0 . Choose a curve T ( λ ) T ( λ ) T(lambda)\mathscr{T}(\lambda)T(λ) whose tangent vector at λ = 0 λ = 0 lambda=0\lambda=0λ=0 is desired.
  1. Compute the number f [ P ( 1 ) ] f [ P ( 0 ) ] f [ P ( 1 ) ] f [ P ( 0 ) ] f[P(1)]-f[P(0)]f[\mathscr{P}(1)]-f[\mathscr{P}(0)]f[P(1)]f[P(0)], which measures the change in f f fff as the point P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) moves from P 0 = P ( 0 ) P 0 = P ( 0 ) P_(0)=P(0)\mathscr{P}_{0}=\mathscr{P}(0)P0=P(0) to Q 0 = P ( 1 ) Q 0 = P ( 1 ) Q_(0)=P(1)\mathscr{Q}_{0}=\mathscr{P}(1)Q0=P(1).
  2. Compute 2 { f [ P ( 1 2 ) ] f [ P ( 0 ) ] } 2 f P 1 2 f [ P ( 0 ) ] 2{f[P((1)/(2))]-f[P(0)]}2\left\{f\left[\mathscr{P}\left(\frac{1}{2}\right)\right]-f[\mathscr{P}(0)]\right\}2{f[P(12)]f[P(0)]}, which is twice the change in f f fff as the point goes from P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0) to P ( 1 2 ) P 1 2 P((1)/(2))\mathscr{P}\left(\frac{1}{2}\right)P(12).
    N N NNN. Compute N { f [ P ( 1 / N ) ] f [ P ( 0 ) ] } N { f [ P ( 1 / N ) ] f [ P ( 0 ) ] } N{f[P(1//N)]-f[P(0)]}N\{f[\mathscr{P}(1 / N)]-f[\mathscr{P}(0)]\}N{f[P(1/N)]f[P(0)]}, which is N N NNN times the change in f f fff as the point goes from P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0) to P ( 1 / N ) P ( 1 / N ) P(1//N)\mathscr{P}(1 / N)P(1/N).
    oo\infty. Same in the limit as N N N longrightarrow ooN \longrightarrow \inftyN : (change in f f fff ) = d f / d λ = d f / d λ =df//d lambda=d f / d \lambda=df/dλ.
    0 . The vector is not itself the change in f f fff. It is instead the operation d / d λ d / d λ d//d lambdad / d \lambdad/dλ, which, when applied to f f fff, gives the change d f / d λ d f / d λ df//d lambdad f / d \lambdadf/dλ. Thus
tangent vector = d / d λ  tangent vector  = d / d λ " tangent vector "=d//d lambda\text { tangent vector }=d / d \lambda tangent vector =d/dλ
[cf. definition (9.1)].
The operation d / d λ d / d λ d//d lambdad / d \lambdad/dλ clearly involves nothing but the last steps N N N longrightarrow ooN \longrightarrow \inftyN in this limiting process, and only those aspects of these steps that are independent of f f fff. But this means it involves the infinitesimal displacements of the point P P P\mathscr{P}P and nothing more.
One who wishes both to stay in touch with the present and to not abandon Cartan's deep geometric insight (Box 9.1) can seek to keep alive a distinction between:
(A) the tangent vector itself in the sense of Cartan, the displacement d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ of a point; and
(B) the "tangent vector operator," or "directional derivative operator," telling what happens to a function in this displacement: (tangent vector operator) = d / d λ = d / d λ =d//d lambda=d / d \lambda=d/dλ.
However, present practice drops (or, if one will, "slurs") the word "operator" in (B), and uses the phrase "tangent vector" itself for the operator, as will be the practice here from now on. The ideas (A) and (B) should also slur or coalesce in one's mind, so that when one visualizes an embedding diagram with arrows drawn tangent to the surface, one always realizes that the arrow characterizes an infinitesimal motion of a point d P / d λ d P / d λ dP//d lambdad \mathscr{P} / d \lambdadP/dλ that takes place purely within the surface, and when one thinks of a derivative operator d / d λ d / d λ d//d lambdad / d \lambdad/dλ, one always visualizes this same infinitesimal motion of a point in the manifold, a motion that must occur in constructing any derivative d f ( P ) / d λ d f ( P ) / d λ df(P)//d lambdad f(\mathscr{P}) / d \lambdadf(P)/dλ. In this sense, one should regard a vector d P / d λ d / d λ d P / d λ d / d λ dP//d lambda-=d//d lambdad \mathscr{P} / d \lambda \equiv d / d \lambdadP/dλd/dλ as both "a displacement that carries attention from one point to another" and "a purely geometric object built on points and nothing but points."
The hard-nosed physicist may still be inclined to say "Tangent vector equals directional derivative operator? Preposterous!" Perhaps he will be put at ease by another argument. He is asked to pick an event P 0 P 0 P_(0)\mathscr{P}_{0}P0. At that event he chooses any set of four noncoplanar vectors (vectors defined in whatever way seems reasonable to him); he names them e 0 , e 1 , e 2 , e 3 e 0 , e 1 , e 2 , e 3 e_(0),e_(1),e_(2),e_(3)\boldsymbol{e}_{0}, \boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e0,e1,e2,e3; and he uses them as a basis on which to expand all other vectors at P 0 P 0 P_(0)\mathscr{P}_{0}P0 :
(9.2) u = u α e α , v = v α e α (9.2) u = u α e α , v = v α e α {:(9.2)u=u^(alpha)e_(alpha)","quad v=v^(alpha)e_(alpha):}\begin{equation*} \boldsymbol{u}=u^{\alpha} \boldsymbol{e}_{\alpha}, \quad \boldsymbol{v}=v^{\alpha} \boldsymbol{e}_{\alpha} \tag{9.2} \end{equation*}(9.2)u=uαeα,v=vαeα
He is asked to construct the four directional derivative operators α e α α e α del_(alpha)-=del_(e_(alpha))\partial_{\alpha} \equiv \partial_{\boldsymbol{e}_{\alpha}}αeα along his four basis vectors. As in flat spacetime, so also here; the same expansion coefficients that appear in u = u α e α u = u α e α u=u^(alpha)e_(alpha)\boldsymbol{u}=u^{\alpha} \boldsymbol{e}_{\alpha}u=uαeα also appear in the expansion for the directional derivative:
(9.3) u = u α α , v = v α α (9.3) u = u α α , v = v α α {:(9.3)del_(u)=u^(alpha)del_(alpha)","quaddel_(v)=v^(alpha)del_(alpha):}\begin{equation*} \partial_{u}=u^{\alpha} \partial_{\alpha}, \quad \partial_{\boldsymbol{v}}=v^{\alpha} \partial_{\alpha} \tag{9.3} \end{equation*}(9.3)u=uαα,v=vαα
Isomorphism between directional derivatives and vectors
Hence, every relation between specific vectors at P 0 P 0 P_(0)\mathscr{P}_{0}P0 induces an identical relation between their differential operators:
(9.4) u = a w + b v u α = a w α + b v α u = a w + b v . (9.4) u = a w + b v u α = a w α + b v α u = a w + b v . {:[(9.4)u=aw+bv<=>u^(alpha)=aw^(alpha)+bv^(alpha)],[ Longleftrightarrowdel_(u)=adel_(w)+bdel_(v).]:}\begin{align*} \boldsymbol{u}=a \boldsymbol{w}+b \boldsymbol{v} & \Leftrightarrow u^{\alpha}=a w^{\alpha}+b v^{\alpha} \tag{9.4}\\ & \Longleftrightarrow \partial_{\boldsymbol{u}}=a \partial_{\boldsymbol{w}}+b \partial_{\mathbf{v}} . \end{align*}(9.4)u=aw+bvuα=awα+bvαu=aw+bv.
There is a complete "isomorphism" between the vectors and the corresponding directional derivatives. So how can the hard-nosed physicist deny the hard-nosed mathematician the right to identify completely each tangent vector with its directional derivative? No harm is done; no answer to any computation can be affected.
This isomorphism extends to the concept "tangent space." Because linear relations (such as u = a w + b v u = a w + b v del_(u)=adel_(w)+bdel_(v)\partial_{\boldsymbol{u}}=a \partial_{\boldsymbol{w}}+b \partial_{\boldsymbol{v}}u=aw+bv ) among directional derivatives evaluated at one and the same point P 0 P 0 P_(0)\mathscr{P}_{0}P0 are meaningful and obey the usual addition and multiplication rules,
Tangent space defined
Coordinate-induced basis defined these derivative operators form an abstract (but finite-dimensional) vector space called the tangent space at P 0 P 0 P_(0)\mathscr{P}_{0}P0. In an embedding picture (Box 9.1) one uses these derivatives (as operators in the flat embedding space) to construct tangent vectors u = u P , v = v P u = u P , v = v P u=del_(u^(P)),v=del_(v)P\boldsymbol{u}=\partial_{\boldsymbol{u}^{\mathscr{P}}}, \boldsymbol{v}=\partial_{\boldsymbol{v}} \mathscr{P}u=uP,v=vP, in the form of straight arrows. Thereby one identifies the abstract tangent space with the geometrically visualized tangent space.

§9.3. BASES, COMPONENTS, AND TRANSFORMATION LAWS FOR VECTORS

An especially useful basis in the tangent space at an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 is induced by any coordinate system [four functions, x 0 ( P ) , x 1 ( P ) , x 2 ( P ) , x 3 ( P ) x 0 ( P ) , x 1 ( P ) , x 2 ( P ) , x 3 ( P ) x^(0)(P),x^(1)(P),x^(2)(P),x^(3)(P)x^{0}(\mathscr{P}), x^{1}(\mathscr{P}), x^{2}(\mathscr{P}), x^{3}(\mathscr{P})x0(P),x1(P),x2(P),x3(P) ]:
(9.5) e 0 ( x 0 ) x 1 , x 2 , x 3 = ( directional derivative along the curve with constant ( x 1 , x 2 , x 3 ) and with parameter λ = x 0 ) at P 0 , e 1 = x 1 , e 2 = x 2 , e 3 = x 3 . (9.5) e 0 x 0 x 1 , x 2 , x 3 =  directional derivative along the   curve with constant  x 1 , x 2 , x 3  and with parameter  λ = x 0 at  P 0 , e 1 = x 1 , e 2 = x 2 , e 3 = x 3 . {:[(9.5)e_(0)-=((del)/(delx^(0)))_(x^(1),x^(2),x^(3))=([" directional derivative along the "],[" curve with constant "(x^(1),x^(2),x^(3))],[" and with parameter "lambda=x^(0)])_("at "P_(0),)],[e_(1)=(del)/(delx^(1))","quade_(2)=(del)/(delx^(2))","quade_(3)=(del)/(delx^(3)).]:}\begin{align*} & \boldsymbol{e}_{0} \equiv\left(\frac{\partial}{\partial x^{0}}\right)_{x^{1}, x^{2}, x^{3}}=\left(\begin{array}{l} \text { directional derivative along the } \\ \text { curve with constant }\left(x^{1}, x^{2}, x^{3}\right) \\ \text { and with parameter } \lambda=x^{0} \end{array}\right)_{\text {at } \mathscr{P}_{0},} \tag{9.5}\\ & \boldsymbol{e}_{1}=\frac{\partial}{\partial x^{1}}, \quad \boldsymbol{e}_{2}=\frac{\partial}{\partial x^{2}}, \quad \boldsymbol{e}_{3}=\frac{\partial}{\partial x^{3}} . \end{align*}(9.5)e0(x0)x1,x2,x3=( directional derivative along the  curve with constant (x1,x2,x3) and with parameter λ=x0)at P0,e1=x1,e2=x2,e3=x3.
(See Figure 9.1.)
A transformation from one basis to another in the tangent space at P 0 P 0 P_(0)\mathscr{\mathscr { P }}_{0}P0, like any change of basis in any vector space, is produced by a nonsingular matrix,
(9.6) e α = e β L β α (9.6) e α = e β L β α {:(9.6)e_(alpha^('))=e_(beta)L^(beta)_(alpha^(')):}\begin{equation*} \boldsymbol{e}_{\alpha^{\prime}}=\boldsymbol{e}_{\beta} L^{\beta}{ }_{\alpha^{\prime}} \tag{9.6} \end{equation*}(9.6)eα=eβLβα
and, as always (including the Lorentz frames of flat spacetime), the components of a vector must transform by the inverse matrix
(9.7) u α = L α β u β ; (9.8) L α β = L β γ 1 , i.e., { L α β L β γ = δ α γ , L δ α L α β = δ δ β . (9.7) u α = L α β u β ; (9.8) L α β = L β γ 1 , i.e.,  L α β L β γ = δ α γ , L δ α L α β = δ δ β . {:[(9.7)u^(alpha^('))=L^(alpha)_(beta)u^(beta);],[(9.8)||L^(alpha^('))_(beta)||=||L^(beta)_(gamma^('))||^(-1)", i.e., "{[L^(alpha^('))_(beta)L^(beta)_(gamma^('))=delta^(alpha^('))_(gamma^('))","],[L^(delta)_(alpha^('))L^(alpha^('))_(beta)=delta^(delta)_(beta).]:}]:}\begin{gather*} u^{\alpha^{\prime}}=L^{\alpha}{ }_{\beta} u^{\beta} ; \tag{9.7}\\ \left\|L^{\alpha^{\prime}}{ }_{\beta}\right\|=\left\|L^{\beta}{ }_{\gamma^{\prime}}\right\|^{-1} \text {, i.e., }\left\{\begin{array}{l} L^{\alpha^{\prime}}{ }_{\beta} L^{\beta}{ }_{\gamma^{\prime}}=\delta^{\alpha^{\prime}}{ }_{\gamma^{\prime}}, \\ L^{\delta}{ }_{\alpha^{\prime}} L^{\alpha^{\prime}}{ }_{\beta}=\delta^{\delta}{ }_{\beta} . \end{array}\right. \tag{9.8} \end{gather*}(9.7)uα=Lαβuβ;(9.8)Lαβ=Lβγ1, i.e., {LαβLβγ=δαγ,LδαLαβ=δδβ.
Figure 9.1.
The basis vectors induced, by a coordinate system, into the tangent space at each event. Here a truncated, two-dimensional spacetime is shown (two other dimensions suppressed), with coordinates χ ( P ) χ ( P ) chi(P)\chi(\mathscr{P})χ(P) and ψ ( P ) ψ ( P ) psi(P)\psi(\mathscr{P})ψ(P), and with corresponding basis vectors / χ / χ del//del chi\partial / \partial \chi/χ and / ψ / ψ del//del psi\partial / \partial \psi/ψ.
This "inverse" transformation law guarantees compatibility between the expansions u = e α u α u = e α u α u=e_(alpha^('))u^(alpha^('))\boldsymbol{u}=\boldsymbol{e}_{\alpha^{\prime}} u^{\alpha^{\prime}}u=eαuα and u = e β u β : u = e β u β : u=e_(beta)u^(beta):\boldsymbol{u}=\boldsymbol{e}_{\beta} u^{\beta}:u=eβuβ:
u = e α u α = ( e γ L γ α ) ( L α β u β ) = e γ δ γ β u β = e β u β . u = e α u α = e γ L γ α L α β u β = e γ δ γ β u β = e β u β . {:[u=e_(alpha^('))u^(alpha^('))=(e_(gamma)L^(gamma)_(alpha))(L^(alpha^('))_(beta)u^(beta))=e_(gamma)delta^(gamma)_(beta)u^(beta)],[=e_(beta)u^(beta).]:}\begin{aligned} \boldsymbol{u} & =\boldsymbol{e}_{\alpha^{\prime}} u^{\alpha^{\prime}}=\left(\boldsymbol{e}_{\gamma} L^{\gamma}{ }_{\alpha}\right)\left(L^{\alpha^{\prime}}{ }_{\beta} u^{\beta}\right)=\boldsymbol{e}_{\gamma} \delta^{\gamma}{ }_{\beta} u^{\beta} \\ & =\boldsymbol{e}_{\beta} u^{\beta} . \end{aligned}u=eαuα=(eγLγα)(Lαβuβ)=eγδγβuβ=eβuβ.
In the special case of transformations between coordinate-induced bases, the transformation matrix has a simple form:
x α = x β x α x β (by usual rules of calculus), x α = x β x α x β  (by usual rules of calculus),  (del)/(delx^(alpha^(')))=(delx^(beta))/(delx^(alpha^(')))(del)/(delx^(beta))" (by usual rules of calculus), "\frac{\partial}{\partial x^{\alpha^{\prime}}}=\frac{\partial x^{\beta}}{\partial x^{\alpha^{\prime}}} \frac{\partial}{\partial x^{\beta}} \text { (by usual rules of calculus), }xα=xβxαxβ (by usual rules of calculus), 
so
(9.9) L β α = ( x β / x α α ) at event Φ 0 where tangent space lies. (9.9) L β α = x β / x α α at event  Φ 0  where tangent space lies.  {:(9.9)L^(beta)_(alpha^('))=(delx^(beta)//delx^(alpha^(alpha)))_("at event "Phi_(0)" where tangent space lies. "):}\begin{equation*} L^{\beta}{ }_{\alpha^{\prime}}=\left(\partial x^{\beta} / \partial x^{\alpha^{\alpha}}\right)_{\text {at event } \Phi_{0} \text { where tangent space lies. }} \tag{9.9} \end{equation*}(9.9)Lβα=(xβ/xαα)at event Φ0 where tangent space lies. 
(Note: this generalizes the Lorentz-transformation law x β = Λ β α x α x β = Λ β α x α x^(beta)=Lambda^(beta)_(alpha)x^(alpha^('))x^{\beta}=\Lambda^{\beta}{ }_{\alpha} x^{\alpha^{\prime}}xβ=Λβαxα, which has the differential form Λ β α = x β / x α Λ β α = x β / x α Lambda^(beta)_(alpha^('))=delx^(beta)//delx^(alpha^('))\Lambda^{\beta}{ }_{\alpha^{\prime}}=\partial x^{\beta} / \partial x^{\alpha^{\prime}}Λβα=xβ/xα; also, it provides a good way to remember the signs in the Λ Λ Lambda\LambdaΛ matrices.)

§9.4. 1-FORMS

When the Lorentz metric is removed from spacetime, one must sharpen up the concept of a 1 -form σ σ sigma\boldsymbol{\sigma}σ by insisting that it, like any tangent vector u u u\boldsymbol{u}u, be attached to a specific event P 0 P 0 P_(0)\mathscr{P}_{0}P0 in spacetime. The family of surfaces representing σ σ sigma\sigmaσ resides in the tangent space at P 0 P 0 P_(0)\mathscr{P}_{0}P0, not in spacetime itself. The piercing of surfaces of σ σ sigma\sigmaσ by an arrow u u u\boldsymbol{u}u to produce the number σ , u σ , u (:sigma,u:)\langle\boldsymbol{\sigma}, \boldsymbol{u}\rangleσ,u ("bongs of bell") occurs entirely in the tangent space.
Figure 9.2.
The basis vectors e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα and dual basis 1-forms ω β ω β omega^(beta)\boldsymbol{\omega}^{\beta}ωβ in the tangent space of an event T 0 T 0 T_(0)\mathscr{T}_{0}T0. The condition
ω β , e α = δ β α ω β , e α = δ β α (:omega^(beta),e_(alpha):)=delta^(beta)_(alpha)\left\langle\boldsymbol{\omega}^{\beta}, \boldsymbol{e}_{\alpha}\right\rangle=\delta^{\beta}{ }_{\alpha}ωβ,eα=δβα
dictates that the vectors e 2 e 2 e_(2)\boldsymbol{e}_{2}e2 and e 3 e 3 e_(3)\boldsymbol{e}_{3}e3 lie parallel to the surfaces of ω 1 ω 1 omega^(1)\boldsymbol{\omega}^{1}ω1, and that e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 extend from one surface of ω 1 ω 1 omega^(1)\boldsymbol{\omega}^{1}ω1 to the next (precisely 1.00 surfaces pierced).
Notice that this picture could fit perfectly well into a book on X-rays and crystallography. There the vectors e 1 , e 2 , e 3 e 1 , e 2 , e 3 e_(1),e_(2),e_(3)\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e1,e2,e3 would be the edges of a unit cell of the crystal; and the surfaces of ω 1 , ω 2 , ω 3 ω 1 , ω 2 , ω 3 omega^(1),omega^(2),omega^(3)\boldsymbol{\omega}^{1}, \boldsymbol{\omega}^{2}, \boldsymbol{\omega}^{3}ω1,ω2,ω3 would be the surfaces of unit cells. Also, for an X-ray diffraction experiment, with wavelength of radiation and orientation of crystal appropriately adjusted, the successive surfaces of ω 1 ω 1 omega^(1)\boldsymbol{\omega}^{1}ω1 would produce Bragg reflection. For other choices of wavelength and orientation, the surfaces of ω 2 ω 2 omega^(2)\boldsymbol{\omega}^{2}ω2 or ω 3 ω 3 omega^(3)\boldsymbol{\omega}^{3}ω3 would produce Bragg reflection.
Given any set of basis vectors { e 0 , e 1 , e 2 , e 3 } e 0 , e 1 , e 2 , e 3 {e_(0),e_(1),e_(2),e_(3)}\left\{\boldsymbol{e}_{0}, \boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}\right\}{e0,e1,e2,e3} at an event P 0 P 0 P_(0)\mathscr{P}_{0}P0, one constructs the
Dual basis of 1 -forms defined
Component-manipulation formulas "dual basis" of 1-forms { ω 0 , ω 1 , ω 2 , ω 3 } ω 0 , ω 1 , ω 2 , ω 3 {omega^(0),omega^(1),omega^(2),omega^(3)}\left\{\boldsymbol{\omega}^{0}, \boldsymbol{\omega}^{1}, \boldsymbol{\omega}^{2}, \boldsymbol{\omega}^{3}\right\}{ω0,ω1,ω2,ω3} by choosing the surfaces of ω β ω β omega^(beta)\boldsymbol{\omega}^{\beta}ωβ such that that
(9.10) ω β , e α = δ β α . (9.10) ω β , e α = δ β α . {:(9.10)(:omega^(beta),e_(alpha):)=delta^(beta)_(alpha).:}\begin{equation*} \left\langle\boldsymbol{\omega}^{\beta}, \boldsymbol{e}_{\alpha}\right\rangle=\delta^{\beta}{ }_{\alpha} . \tag{9.10} \end{equation*}(9.10)ωβ,eα=δβα.
See Figure 9.2. A marvelously simple formalism for calculating and manipulating components of tangent vectors and 1 -forms then results:
(9.11a) u = e α u α ( definition of components of u ) , (9.11b) σ = σ β ω β (definition of components of σ ), (9.11c) u α = ω α , u (way to calculate components of u ), (9.11d) σ β = σ , e β (way to calculate components of σ ), (9.11e) σ , u = σ α u α (way to calculate σ , u using components), (9.11f) ω α = L α β β (transformation law for 1-form basis, corresponding to equation 9.6), (9.11~g) σ α = σ β L β α (transformation law for 1-form components). (9.11a) u = e α u α (  definition of components of  u ) , (9.11b) σ = σ β ω β  (definition of components of  σ  ),  (9.11c) u α = ω α , u  (way to calculate components of  u  ),  (9.11d) σ β = σ , e β  (way to calculate components of  σ  ),  (9.11e) σ , u = σ α u α  (way to calculate  σ , u  using   components),  (9.11f) ω α = L α β β  (transformation law for 1-form   basis, corresponding to equation 9.6),  (9.11~g) σ α = σ β L β α  (transformation law for 1-form   components).  {:[(9.11a)u=e_(alpha)u^(alpha)quad(" definition of components of "u)","],[(9.11b)sigma=sigma_(beta)omega^(beta)quad" (definition of components of "sigma" ), "],[(9.11c)u^(alpha)=(:omega^(alpha),u:)quad" (way to calculate components of "u" ), "],[(9.11d)sigma_(beta)=(:sigma,e_(beta):)quad" (way to calculate components of "sigma" ), "],[(9.11e)(:sigma","u:)=sigma_(alpha)u^(alpha)quad" (way to calculate "(:sigma","u:)" using "],[" components), "],[(9.11f)omega^(alpha^('))=L^(alpha^('))beta^(beta)quad" (transformation law for 1-form "],[" basis, corresponding to equation 9.6), "],[(9.11~g)sigma_(alpha^('))=sigma_(beta)L^(beta)_(alpha)quad" (transformation law for 1-form "],[" components). "]:}\begin{align*} & \boldsymbol{u}=\boldsymbol{e}_{\alpha} u^{\alpha} \quad(\text { definition of components of } \boldsymbol{u}), \tag{9.11a}\\ & \boldsymbol{\sigma}=\sigma_{\beta} \boldsymbol{\omega}^{\beta} \quad \text { (definition of components of } \boldsymbol{\sigma} \text { ), } \tag{9.11b}\\ & u^{\alpha}=\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{u}\right\rangle \quad \text { (way to calculate components of } \boldsymbol{u} \text { ), } \tag{9.11c}\\ & \sigma_{\beta}=\left\langle\boldsymbol{\sigma}, \boldsymbol{e}_{\beta}\right\rangle \quad \text { (way to calculate components of } \boldsymbol{\sigma} \text { ), } \tag{9.11d}\\ & \langle\boldsymbol{\sigma}, \boldsymbol{u}\rangle=\sigma_{\alpha} u^{\alpha} \quad \text { (way to calculate }\langle\boldsymbol{\sigma}, \boldsymbol{u}\rangle \text { using } \tag{9.11e}\\ & \text { components), } \\ & \boldsymbol{\omega}^{\alpha^{\prime}}=L^{\alpha^{\prime}} \boldsymbol{\beta}^{\beta} \quad \text { (transformation law for 1-form } \tag{9.11f}\\ & \text { basis, corresponding to equation 9.6), } \\ & \sigma_{\alpha^{\prime}}=\sigma_{\beta} L^{\beta}{ }_{\alpha} \quad \text { (transformation law for 1-form } \tag{9.11~g}\\ & \text { components). } \end{align*}(9.11a)u=eαuα( definition of components of u),(9.11b)σ=σβωβ (definition of components of σ ), (9.11c)uα=ωα,u (way to calculate components of u ), (9.11d)σβ=σ,eβ (way to calculate components of σ ), (9.11e)σ,u=σαuα (way to calculate σ,u using  components), (9.11f)ωα=Lαββ (transformation law for 1-form  basis, corresponding to equation 9.6), (9.11~g)σα=σβLβα (transformation law for 1-form  components). 
(Exercise 9.1 below justifies these equations.)
In the absence of a metric, there is no way to pick a specific 1 -form u ~ u ~ widetilde(u)\widetilde{\boldsymbol{u}}u~ at an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 and say that it corresponds to a specific tangent vector u u u\boldsymbol{u}u at P 0 P 0 P_(0)\mathscr{P}_{0}P0. The correspondence set up in flat spacetime,
u ~ , v = u v for all v u ~ , v = u v  for all  v (: widetilde(u),v:)=u*v quad" for all "v\langle\widetilde{\boldsymbol{u}}, \boldsymbol{v}\rangle=\boldsymbol{u} \cdot \boldsymbol{v} \quad \text { for all } \boldsymbol{v}u~,v=uv for all v
was rubbed out when "•" was rubbed out. Restated in component language: the raising of an index, u α = η α β u β u α = η α β u β u^(alpha)=eta^(alpha beta)u_(beta)u^{\alpha}=\eta^{\alpha \beta} u_{\beta}uα=ηαβuβ, is impossible because the η α β η α β eta^(alpha beta)\eta^{\alpha \beta}ηαβ do not exist; similarly, lowering of an index, u β = η β α u α u β = η β α u α u_(beta)=eta_(beta alpha)u^(alpha)u_{\beta}=\eta_{\beta \alpha} u^{\alpha}uβ=ηβαuα, is impossible.
The 1 -form gradient d f d f df\boldsymbol{d} fdf was introduced in § 2.6 § 2.6 §2.6\S 2.6§2.6 with absolutely no reference to metric. Consequently, it and its mathematical formalism are the same here, without metric, as there with metric, except that, like all other 1 -forms, d f d f df\boldsymbol{d} fdf now resides in the tangent space rather than in spacetime itself. For example, there is no change in the fundamental equation relating the projection of the gradient to the directional derivative:
(9.12) d f , u = u f = u [ f ] . [ old notation for directional derivative ] new notation; recall u = u ] . (9.12) d f , u = u f = u [ f ] .  old notation for   directional derivative   new notation;   recall  u = u . {:(9.12){:[(:df","u:)=del_(u)f=u[f].],[[[" old notation for "],[" directional derivative "]]]quad[" new notation; "],[" recall "u=del_(u)]].:}\left.\begin{array}{c} \langle\boldsymbol{d} f, \boldsymbol{u}\rangle=\partial_{\boldsymbol{u}} f=\boldsymbol{u}[f] . \tag{9.12}\\ {\left[\begin{array}{r} \text { old notation for } \\ \text { directional derivative } \end{array}\right]} \end{array} \quad \begin{array}{l} \text { new notation; } \\ \text { recall } \boldsymbol{u}=\partial_{\boldsymbol{u}} \end{array}\right] .(9.12)df,u=uf=u[f].[ old notation for  directional derivative ] new notation;  recall u=u].
Similarly, there are no changes in the component equations,
(家) d f = f , α ω α (expansion of d f in arbitrary basis), f , α = α f = e α [ f ] (way to calculate components of d f ), f , α = f / x α if { e α } is a coordinate basis, (家) d f = f , α ω α  (expansion of  d f  in arbitrary   basis),  f , α = α f = e α [ f ]  (way to calculate components   of  d f  ),  f , α = f / x α  if  e α  is a coordinate basis,  {:(家){:[df=f_(,alpha)omega^(alpha),{:[" (expansion of "df" in arbitrary "],[" basis), "]:}],[f_(,alpha)=del_(alpha)f=e_(alpha)[f],{:[" (way to calculate components "],[" of "df" ), "]:}],[f_(,alpha)=del f//delx^(alpha)," if "{e_(alpha)}" is a coordinate basis, "]:}:}\begin{array}{ll} \boldsymbol{d} f=f_{, \alpha} \boldsymbol{\omega}^{\alpha} & \begin{array}{l} \text { (expansion of } \boldsymbol{d} f \text { in arbitrary } \\ \text { basis), } \end{array} \\ f_{, \alpha}=\partial_{\alpha} f=\boldsymbol{e}_{\alpha}[f] & \begin{array}{l} \text { (way to calculate components } \\ \text { of } \boldsymbol{d} f \text { ), } \end{array} \tag{家}\\ f_{, \alpha}=\partial f / \partial x^{\alpha} & \text { if }\left\{\boldsymbol{e}_{\alpha}\right\} \text { is a coordinate basis, } \end{array}(家)df=f,αωα (expansion of df in arbitrary  basis), f,α=αf=eα[f] (way to calculate components  of df ), f,α=f/xα if {eα} is a coordinate basis, 
except that they work in arbitrary bases, not just in Lorentz bases. And, as in Lorentz frames, so also in general: the one-form basis { d x α } d x α {dx^(alpha)}\left\{\boldsymbol{d} x^{\alpha}\right\}{dxα} and the tangent-vector basis { / x α } / x α {del//delx^(alpha)}\left\{\partial / \partial x^{\alpha}\right\}{/xα}, which are induced into a tangent space by the same coordinate system, are the duals of each other,
(9.14) d x α , / x β = δ α β (9.14) d x α , / x β = δ α β {:(9.14)(:dx^(alpha),del//delx^(beta):)=delta^(alpha)_(beta):}\begin{equation*} \left\langle\boldsymbol{d} x^{\alpha}, \partial / \partial x^{\beta}\right\rangle=\delta^{\alpha}{ }_{\beta} \tag{9.14} \end{equation*}(9.14)dxα,/xβ=δαβ
(See exercise 9.2 for proofs.) Also, most aspects of Cartan's "Exterior Calculus" (parts A, B, C of Box 4.1) are left unaffected by the removal of metric.

§9.5. TENSORS

A tensor S S S\boldsymbol{S}S, in the absence of Lorentz metric, differs from the tensors of flat, Lorentz spacetime in two ways. (1) S S S\boldsymbol{S}S must reside at a specific event P 0 P 0 P_(0)\mathscr{P}_{0}P0, just as any vector or 1-form must. (2) Each slot of S S S\boldsymbol{S}S is specific; it will accept either vectors or 1 -forms, but not both, because it has no way to convert a 1-form u ~ u ~ widetilde(u)\widetilde{\boldsymbol{u}}u~ into a "corresponding
Similarly
Correspondence between vectors and 1 -forms rubbed out
Gradient of a function ?
vector" u u u\boldsymbol{u}u as it sends u ~ u ~ widetilde(u)\widetilde{\boldsymbol{u}}u~ through its linear machinery. Thus, if S S S\boldsymbol{S}S is a ( 1 2 ) 1 2 ((1)/(2))\left(\frac{1}{2}\right)(12) tensor

then it cannot be converted alternatively to a ( 2 1 ) ( 2 1 ) ((2)/(1))\binom{2}{1}(21) tensor, or a ( 3 0 ) ( 3 0 ) ((3)/(0))\binom{3}{0}(30) tensor, or a ( 0 3 ) ( 0 3 ) ((0)/(3))\binom{0}{3}(03) tensor by the procedure of § 3.2 § 3.2 §3.2\S 3.2§3.2. In component language, the indices of S S S\boldsymbol{S}S cannot be raised and lowered.
Except for these two restrictions (attachment to a specific event; specificity of slots), a tensor S S S\boldsymbol{S}S is the same linear machine as ever. And the algebra of component manipulations is the same:
(9.16) S α β γ = S ( ω α , e β , e γ ) ( S , ω α , e β must all reside at same event ) (9.17) S = S α β γ e α ω β ω γ , (9.18) S ( σ , u , v ) = S α β γ σ α u β v γ . (9.16) S α β γ = S ω α , e β , e γ S , ω α , e β  must all reside at same event  (9.17) S = S α β γ e α ω β ω γ , (9.18) S ( σ , u , v ) = S α β γ σ α u β v γ . {:[(9.16)S^(alpha)_(beta gamma)=S(omega^(alpha),e_(beta),e_(gamma))quad(S,omega^(alpha),e_(beta)" must all reside at same event ")],[(9.17)S=S^(alpha)_(beta gamma)e_(alpha)oxomega^(beta)oxomega^(gamma)","],[(9.18)S(sigma","u","v)=S^(alpha)_(beta gamma)^(sigma)_(alpha)u^(beta)v^(gamma).]:}\begin{align*} S^{\alpha}{ }_{\beta \gamma} & =\boldsymbol{S}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}\right) \quad\left(\boldsymbol{S}, \boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta} \text { must all reside at same event }\right) \tag{9.16}\\ \boldsymbol{S} & =S^{\alpha}{ }_{\beta \gamma} \boldsymbol{e}_{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma}, \tag{9.17}\\ \boldsymbol{S}(\boldsymbol{\sigma}, \boldsymbol{u}, \boldsymbol{v}) & =S^{\alpha}{ }_{\beta \gamma}{ }^{\sigma}{ }_{\alpha} u^{\beta} v^{\gamma} . \tag{9.18} \end{align*}(9.16)Sαβγ=S(ωα,eβ,eγ)(S,ωα,eβ must all reside at same event )(9.17)S=Sαβγeαωβωγ,(9.18)S(σ,u,v)=Sαβγσαuβvγ.

EXERCISES

Exercise 9.1. COMPONENT MANIPULATIONS

Derive equations ( 9.11 c ) through ( 9.11 g ( 9.11 g (9.11g(9.11 \mathrm{~g}(9.11 g ) from (9.10), (9.11a, b), (9.6), (9.7), and (9.8).

Exercise 9.2. COMPONENTS OF GRADIENT, AND DUALITY OF COORDINATE BASES

In an arbitrary basis, define f , α f , α f_(,alpha)f_{, \alpha}f,α by the expansion (9.13a). Then combine equations ( 9.11 d ) and (9.12) to obtain the method (9.13b) of computing f , α f , α f_(,alpha)f_{, \alpha}f,α. Finally, combine equations (9.12) and ( 9.13 b ) ( 9.13 b ) (9.13b)(9.13 \mathrm{~b})(9.13 b) to show that the bases { d x α } d x α {dx^(alpha)}\left\{\boldsymbol{d} x^{\alpha}\right\}{dxα} and { / x β } / x β {del//delx^(beta)}\left\{\partial / \partial x^{\beta}\right\}{/xβ} are the duals of each other.

Exercise 9.3. PRACTICE MANIPULATING TANGENT VECTORS

Let P 0 P 0 P_(0)\mathscr{P}_{0}P0 be the point with coordinates ( x = 0 , y = 1 , z = 0 ) ( x = 0 , y = 1 , z = 0 ) (x=0,y=1,z=0)(x=0, y=1, z=0)(x=0,y=1,z=0) in a three-dimensional space; and define three curves through P 0 P 0 P_(0)\mathscr{P}_{0}P0 by
P ( λ ) = ( λ , 1 , λ ) P ( ζ ) = ( sin ζ , cos ζ , ζ ) P ( ρ ) = ( sinh ρ , cosh ρ , ρ + ρ 3 ) P ( λ ) = ( λ , 1 , λ ) P ( ζ ) = ( sin ζ , cos ζ , ζ ) P ( ρ ) = sinh ρ , cosh ρ , ρ + ρ 3 {:[P(lambda)=(lambda","1","lambda)],[P(zeta)=(sin zeta","cos zeta","zeta)],[P(rho)=(sinh rho,cosh rho,rho+rho^(3))]:}\begin{aligned} \mathscr{P}(\lambda) & =(\lambda, 1, \lambda) \\ \mathscr{P}(\zeta) & =(\sin \zeta, \cos \zeta, \zeta) \\ \mathscr{P}(\rho) & =\left(\sinh \rho, \cosh \rho, \rho+\rho^{3}\right) \end{aligned}P(λ)=(λ,1,λ)P(ζ)=(sinζ,cosζ,ζ)P(ρ)=(sinhρ,coshρ,ρ+ρ3)
(a) Compute ( d / d λ ) f ( d / d λ ) f (d//d lambda)f(d / d \lambda) f(d/dλ)f, ( d / d ζ ) f ( d / d ζ ) f (d//d zeta)f(d / d \zeta) f(d/dζ)f, and ( d / d ρ ) f ( d / d ρ ) f (d//d rho)f(d / d \rho) f(d/dρ)f for the function f = x 2 y 2 + z 2 f = x 2 y 2 + z 2 f=x^(2)-y^(2)+z^(2)f=x^{2}-y^{2}+z^{2}f=x2y2+z2 at the point P 0 P 0 P_(0)\mathscr{P}_{0}P0. (b) Calculate the components of the tangent vectors d / d λ , d / d ζ d / d λ , d / d ζ d//d lambda,d//d zetad / d \lambda, d / d \zetad/dλ,d/dζ, and d / d ρ d / d ρ d//d rhod / d \rhod/dρ at P 0 P 0 P_(0)\mathscr{P}_{0}P0, using the basis { / x , / y , / z } { / x , / y , / z } {del//del x,del//del y,del//del z}\{\partial / \partial x, \partial / \partial y, \partial / \partial z\}{/x,/y,/z}.

Exercise 9.4. MORE PRACTICE WITH TANGENT VECTORS

In a three-dimensional space with coordinates ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z), introduce the vector field v = y 2 v = y 2 v=y^(2)\boldsymbol{v}=y^{2}v=y2 / x x / z / x x / z del//del x-x del//del z\partial / \partial x-x \partial / \partial z/xx/z, and the functions f = x y , g = z 3 f = x y , g = z 3 f=xy,g=z^(3)f=x y, g=z^{3}f=xy,g=z3. Compute
(a) v [ f ] v [ f ] v[f]\boldsymbol{v}[f]v[f]
(c) v [ f g ] v [ f g ] v[fg]\boldsymbol{v}[f g]v[fg]
(e) v [ f 2 + g 2 ] v f 2 + g 2 v[f^(2)+g^(2)]\boldsymbol{v}\left[f^{2}+g^{2}\right]v[f2+g2]
(b) v [ g ] v [ g ] v[g]\boldsymbol{v}[g]v[g]
(d) f v [ g ] g v [ f ] f v [ g ] g v [ f ] fv[g]-gv[f]f \boldsymbol{v}[g]-g \boldsymbol{v}[f]fv[g]gv[f]
(f) v { v [ f ] } v { v [ f ] } v{v[f]}\boldsymbol{v}\{\boldsymbol{v}[f]\}v{v[f]}

Exercise 9.5. PICTURE OF BASIS 1-FORMS INDUCED BY COORDINATES

In the tangent space of Figure 9.1, draw the basis 1 -forms d ψ d ψ d psi\boldsymbol{d} \psidψ and d χ d χ d_(chi)\boldsymbol{d}_{\chi}dχ induced by the ψ , χ ψ , χ psi,chi\psi, \chiψ,χ-coordinate system.

Exercise 9.6. PRACTICE WITH DUAL BASES

In a three-dimensional space with spherical coordinates r , θ r , θ r,thetar, \thetar,θ, ϕ ϕ phi\phiϕ, one often likes to use, instead of the basis / r , / θ , / ϕ / r , / θ , / ϕ del//del r,del//del theta,del//del phi\partial / \partial r, \partial / \partial \theta, \partial / \partial \phi/r,/θ,/ϕ, the basis
e r ^ = r , e θ ^ = 1 r θ , e ϕ ^ = 1 r sin θ ϕ e r ^ = r , e θ ^ = 1 r θ , e ϕ ^ = 1 r sin θ ϕ e_( hat(r))=(del)/(del r),quade_( hat(theta))=(1)/(r)(del)/(del theta),quade_( hat(phi))=(1)/(r sin theta)(del)/(del phi)\boldsymbol{e}_{\hat{r}}=\frac{\partial}{\partial r}, \quad \boldsymbol{e}_{\hat{\theta}}=\frac{1}{r} \frac{\partial}{\partial \theta}, \quad \boldsymbol{e}_{\hat{\phi}}=\frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}er^=r,eθ^=1rθ,eϕ^=1rsinθϕ
(a) What is the 1 -form basis { ω γ ^ , ω θ ^ , ω ϕ ^ } ω γ ^ , ω θ ^ , ω ϕ ^ {omega^( hat(gamma)),omega^( hat(theta)),omega^( hat(phi))}\left\{\boldsymbol{\omega}^{\hat{\gamma}}, \boldsymbol{\omega}^{\hat{\theta}}, \boldsymbol{\omega}^{\hat{\phi}}\right\}{ωγ^,ωθ^,ωϕ^} dual to this tangent-vector basis? (b) On the sphere r = 1 r = 1 r=1r=1r=1, draw pictures of the bases { / r , / θ , / ϕ } , { e r ^ , e θ ^ , e ϕ ^ } , { d r , d θ , d ϕ } { / r , / θ , / ϕ } , e r ^ , e θ ^ , e ϕ ^ , { d r , d θ , d ϕ } {del//del r,del//del theta,del//del phi},{e_( hat(r)),e_( hat(theta)),e_( hat(phi))},{dr,d theta,d phi}\{\partial / \partial r, \partial / \partial \theta, \partial / \partial \phi\},\left\{\boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}, \boldsymbol{e}_{\hat{\phi}}\right\},\{\boldsymbol{d} r, \boldsymbol{d} \theta, \boldsymbol{d} \phi\}{/r,/θ,/ϕ},{er^,eθ^,eϕ^},{dr,dθ,dϕ}, and { ω r ^ , ω θ ^ , ω ϕ ^ } ω r ^ , ω θ ^ , ω ϕ ^ {omega^( hat(r)),omega^( hat(theta)),omega^( hat(phi))}\left\{\boldsymbol{\omega}^{\hat{r}}, \boldsymbol{\omega}^{\hat{\theta}}, \boldsymbol{\omega}^{\hat{\phi}}\right\}{ωr^,ωθ^,ωϕ^}.

§9.6. COMMUTATORS AND PICTORIAL TECHNIQUES

A vector u 0 u 0 u_(0)\boldsymbol{u}_{0}u0 given only at one point P 0 P 0 P_(0)\mathscr{P}_{0}P0 suffices to compute the derivative u 0 [ f ] u 0 [ f ] u_(0)[f]-=\boldsymbol{u}_{0}[f] \equivu0[f] u 0 f u 0 f del_(u_(0))f\partial_{\boldsymbol{u}_{0}} fu0f, which is simply a number associated with the point P 0 P 0 P_(0)\mathscr{P}_{0}P0. In contrast, a vector field u u u\boldsymbol{u}u provides a vector u ( P ) u ( P ) u(P)\boldsymbol{u}(\mathscr{P})u(P)-which is a differential operator u ( P ) u ( P ) del_(u(P))\partial_{\boldsymbol{u}(\mathscr{P})}u(P)-at each point P P P\mathscr{P}P in some region of spacetime. This vector field operates on a function f f fff to produce not just a number, but another function u [ f ] u f u [ f ] u f u[f]-=del_(u)f\boldsymbol{u}[f] \equiv \partial_{\boldsymbol{u}} fu[f]uf. A second vector field v v v\boldsymbol{v}v can perfectly well operate on this new function, to produce yet another function
v { u [ f ] } = v ( u f ) v { u [ f ] } = v u f v{u[f]}=del_(v)(del_(u)f)\boldsymbol{v}\{\boldsymbol{u}[f]\}=\partial_{\boldsymbol{v}}\left(\partial_{\boldsymbol{u}} f\right)v{u[f]}=v(uf)
Does this function agree with the result of applying v v v\boldsymbol{v}v first and then u u u\boldsymbol{u}u ? Equivalently, does the "commutator"
(9.19) [ u , v ] [ f ] u { v [ f ] } v { u [ f ] } (9.19) [ u , v ] [ f ] u { v [ f ] } v { u [ f ] } {:(9.19)[u","v][f]-=u{v[f]}-v{u[f]}:}\begin{equation*} [\boldsymbol{u}, \boldsymbol{v}][f] \equiv \boldsymbol{u}\{\boldsymbol{v}[f]\}-\boldsymbol{v}\{\boldsymbol{u}[f]\} \tag{9.19} \end{equation*}(9.19)[u,v][f]u{v[f]}v{u[f]}
Commutator defined
vanish?
The simplest special case is when u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are basis vectors of a coordinate system, u = / x α , v = / x β u = / x α , v = / x β u=del//delx^(alpha),v=del//delx^(beta)\boldsymbol{u}=\partial / \partial x^{\alpha}, \boldsymbol{v}=\partial / \partial x^{\beta}u=/xα,v=/xβ. Then the commutator does vanish, because partial derivatives always commute:
[ / x α , / x β ] [ f ] = 2 f / x β x α 2 f / x α x β = 0 / x α , / x β [ f ] = 2 f / x β x α 2 f / x α x β = 0 [del//delx^(alpha),del//delx^(beta)][f]=del^(2)f//delx^(beta)delx^(alpha)-del^(2)f//delx^(alpha)delx^(beta)=0\left[\partial / \partial x^{\alpha}, \partial / \partial x^{\beta}\right][f]=\partial^{2} f / \partial x^{\beta} \partial x^{\alpha}-\partial^{2} f / \partial x^{\alpha} \partial x^{\beta}=0[/xα,/xβ][f]=2f/xβxα2f/xαxβ=0
But in general the commutator is nonzero, as one sees from a coordinate-based calculation:
[ u , v ] f = u α x α ( v β f x β ) v α x α ( u β f x β ) = [ ( u α v β , α v α u β , α ) x β ] f . [ u , v ] f = u α x α v β f x β v α x α u β f x β = u α v β , α v α u β , α x β f . {:[[u","v]f=u^(alpha)(del)/(delx^(alpha))(v^(beta)(del f)/(delx^(beta)))-v^(alpha)(del)/(delx^(alpha))(u^(beta)(del f)/(delx^(beta)))],[=[(u^(alpha)v^(beta)_(,alpha)-v^(alpha)u^(beta)_(,alpha))(del)/(delx^(beta))]f.]:}\begin{aligned} {[\boldsymbol{u}, \boldsymbol{v}] f } & =u^{\alpha} \frac{\partial}{\partial x^{\alpha}}\left(v^{\beta} \frac{\partial f}{\partial x^{\beta}}\right)-v^{\alpha} \frac{\partial}{\partial x^{\alpha}}\left(u^{\beta} \frac{\partial f}{\partial x^{\beta}}\right) \\ & =\left[\left(u^{\alpha} v^{\beta}{ }_{, \alpha}-v^{\alpha} u^{\beta}{ }_{, \alpha}\right) \frac{\partial}{\partial x^{\beta}}\right] f . \end{aligned}[u,v]f=uαxα(vβfxβ)vαxα(uβfxβ)=[(uαvβ,αvαuβ,α)xβ]f.
Commutator of two vector fields is a vector field
Commutator as a "closer of curves" ^('){ }^{\prime}
Notice however, that the commutator [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v], like u u u\boldsymbol{u}u and v v v\boldsymbol{v}v themselves, is a vector field, i.e., a linear differential operator at each event:
(9.20) [ u , v ] = ( u [ v β ] v [ u β ] ) x β = ( u α v β , α v α u β , α ) x β (9.20) [ u , v ] = u v β v u β x β = u α v β , α v α u β , α x β {:(9.20)[u","v]=(u[v^(beta)]-v[u^(beta)])(del)/(delx^(beta))=(u^(alpha)v^(beta)_(,alpha)-v^(alpha)u^(beta)_(,alpha))(del)/(delx^(beta)):}\begin{equation*} [\boldsymbol{u}, \boldsymbol{v}]=\left(\boldsymbol{u}\left[v^{\beta}\right]-\boldsymbol{v}\left[u^{\beta}\right]\right) \frac{\partial}{\partial x^{\beta}}=\left(u^{\alpha} v^{\beta}{ }_{, \alpha}-v^{\alpha} u^{\beta}{ }_{, \alpha}\right) \frac{\partial}{\partial x^{\beta}} \tag{9.20} \end{equation*}(9.20)[u,v]=(u[vβ]v[uβ])xβ=(uαvβ,αvαuβ,α)xβ
Such results should be familiar from quantum theory's formalism for angular momentum operators (exercise 9.8).
The three levels of geometry-pictorial, abstract, and component-yield three different insights into the commutator. (1) The abstract expression [ u , v u , v u,v\boldsymbol{u}, \boldsymbol{v}u,v ] suggests the close connection to quantum theory, and brings to mind the many tools developed there for handling operators. But recall that the operators of quantum theory need not be first-order differential operators. The kinetic energy is second order and the potential is zeroth order in the familiar Schrödinger equation. Only first-order operators are vectors. (2) The component expression u α v β , α v α u β , α u α v β , α v α u β , α u^(alpha)v^(beta)_(,alpha)-v^(alpha)u^(beta)_(,alpha)u^{\alpha} v^{\beta}{ }_{, \alpha}-v^{\alpha} u^{\beta}{ }_{, \alpha}uαvβ,αvαuβ,α, valid in any coordinate basis, brings the commutator into the reaches of the powerful tools of index mechanics. (3) The pictorial representation of [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] (Box 9.2) reveals its fundamental role as a "closer of curves"-a role that will be important in Chapter 11's analysis of curvature.
Commutators find application in the distinction between a coordinate-induced basis, { e α } = { / x α } e α = / x α {e_(alpha)}={del//delx^(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}=\left\{\partial / \partial x^{\alpha}\right\}{eα}={/xα}, and a noncoordinate basis. Because partial derivatives always commute,
(9.21) [ e α , e β ] = [ / x α , / x β ] = 0 in any coordinate basis. (9.21) e α , e β = / x α , / x β = 0  in any coordinate basis.  {:(9.21)[e_(alpha),e_(beta)]=[del//delx^(alpha),del//delx^(beta)]=0" in any coordinate basis. ":}\begin{equation*} \left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]=\left[\partial / \partial x^{\alpha}, \partial / \partial x^{\beta}\right]=0 \text { in any coordinate basis. } \tag{9.21} \end{equation*}(9.21)[eα,eβ]=[/xα,/xβ]=0 in any coordinate basis. 

Box 9.2 THE COMMUTATOR AS A CLOSER OF QUADRILATERALS

A. Pictorial Representation in Flat Spacetime

  1. For ease of visualization, consider flat spacetime, so the two vector fields u ( P ) u ( P ) u(P)\boldsymbol{u}(\mathscr{P})u(P) and v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathscr{P})v(P) can be laid out in spacetime itself.
  2. Choose an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 where the commutator [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] is to be calculated.
  3. Give the names P 1 , P 2 , P 3 , P 4 P 1 , P 2 , P 3 , P 4 P_(1),P_(2),P_(3),P_(4)\mathscr{P}_{1}, \mathscr{P}_{2}, \mathscr{P}_{3}, \mathscr{P}_{4}P1,P2,P3,P4 to the events pictured in the diagram.
  4. Then the vector P 4 P 3 P 4 P 3 P_(4)-P_(3)\mathscr{P}_{4}-\mathscr{P}_{3}P4P3, which measures how much the four-legged curve fails to close, can be expressed in a coordinate basis as
P 4 P 3 = [ u ( P 0 ) + v ( P 1 ) ] [ u ( P 2 ) + v ( P 0 ) ] = [ v ( P 1 ) v ( P 0 ) ] [ u ( P 2 ) u ( P 0 ) ] = ( v β , α u α e β ) Q 0 ( u β , α v α e β ) φ 0 + errors = [ u , v ] P 0 + errors. [terms such as v β , μ ν u μ u ν e β ] P 4 P 3 = u P 0 + v P 1 u P 2 + v P 0 = v P 1 v P 0 u P 2 u P 0 = v β , α u α e β Q 0 u β , α v α e β φ 0 +  errors  = [ u , v ] P 0 +  errors.   [terms such as  v β , μ ν u μ u ν e β  ]  {:[P_(4)-P_(3)=[u(P_(0))+v(P_(1))]-[u(P_(2))+v(P_(0))]],[=[v(P_(1))-v(P_(0))]-[u(P_(2))-u(P_(0))]],[=(v^(beta)_(,alpha)u^(alpha)e_(beta))_(Q_(0))-(u^(beta)_(,alpha)v^(alpha)e_(beta))_(varphi_(0))+" errors "],[=[u","v]_(P_(0))+" errors. "],[" [terms such as "v^(beta)_(,mu nu)u^(mu)u^(nu)e_(beta)" ] "]:}\begin{aligned} & \mathscr{P}_{4}-\mathscr{P}_{3}=\left[\boldsymbol{u}\left(\mathscr{P}_{0}\right)+\boldsymbol{v}\left(\mathscr{P}_{1}\right)\right]-\left[\boldsymbol{u}\left(\mathscr{P}_{2}\right)+\boldsymbol{v}\left(\mathscr{P}_{0}\right)\right] \\ & =\left[\boldsymbol{v}\left(\mathscr{P}_{1}\right)-\boldsymbol{v}\left(\mathscr{P}_{0}\right)\right]-\left[\boldsymbol{u}\left(\mathscr{P}_{2}\right)-\boldsymbol{u}\left(\mathscr{P}_{0}\right)\right] \\ & =\left(v^{\beta}{ }_{, \alpha} u^{\alpha} \boldsymbol{e}_{\beta}\right)_{\mathscr{Q}_{0}}-\left(u^{\beta}{ }_{, \alpha} v^{\alpha} \boldsymbol{e}_{\beta}\right)_{\mathscr{\varphi}_{0}}+\text { errors } \\ & =[\boldsymbol{u}, \boldsymbol{v}]_{\mathscr{P}_{0}}+\text { errors. } \\ & \text { [terms such as } v^{\beta}{ }_{, \mu \nu} u^{\mu} u^{\nu} \boldsymbol{e}_{\beta} \text { ] } \end{aligned}P4P3=[u(P0)+v(P1)][u(P2)+v(P0)]=[v(P1)v(P0)][u(P2)u(P0)]=(vβ,αuαeβ)Q0(uβ,αvαeβ)φ0+ errors =[u,v]P0+ errors.  [terms such as vβ,μνuμuνeβ ] 
  1. Notice that if u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are halved everywhere, then [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] is cut down by a factor of 4 , while the error terms in the above go down by a factor of 8 . Thus, [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] represents accurately the gap in the four-legged curve ("quadrilateral") in the limit where u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are sufficiently short; i.e., [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] "closes the quadrilateral" whose edges are the vectors fields u u u\boldsymbol{u}u and v v v\boldsymbol{v}v.

B. Pictorial Representation in Absence of Metric, or in Curved Spacetime with a Metric

  1. The same picture must work, but now one dares not (at least initially) lay out the vector fields in spacetime itself. Instead one lays out two families of curves: the curves for which u ( P ) u ( P ) u(P)\boldsymbol{u}(\mathscr{P})u(P) is the tangent vector; and the curves for which v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathscr{P})v(P) is the tangent vector.
  2. The gap " P 4 P 3 P 4 P 3 P_(4)-P_(3)\mathscr{P}_{4}-\mathscr{P}_{3}P4P3 " in the four-legged curve can be characterized by the difference f ( P 4 ) f ( P 3 ) f P 4 f P 3 f(P_(4))-f(P_(3))f\left(\mathscr{P}_{4}\right)-f\left(\mathscr{P}_{3}\right)f(P4)f(P3) in the values of an arbitrary function at P 4 P 4 P_(4)\mathscr{P}_{4}P4 and P 3 P 3 P_(3)\mathscr{P}_{3}P3. That difference is, in a coordinate basis,

Box 9.2 (continued)

f ( P 4 ) f ( P 3 ) = [ f ( P 4 ) f ( P 1 ) ] ( f , α v α + 1 2 f , α β v α v β ) P 1 ( , α , u α + 1 2 f , α β u α u β ) P 0 ) f ( P 0 ) ] [ f ( P 2 ) f ( P 0 ) ] [ f ( P 3 ) f ( P 2 ) ] ] f P 4 f P 3 = f P 4 f P 1 f , α v α + 1 2 f , α β v α v β P 1 ( , α , u α + 1 2 f , α β u α u β ) P 0 ) f P 0 ] f P 2 f P 0 f P 3 f P 2 ] {:[f(P_(4))-f(P_(3))=,ubrace([f(P_(4))-f(P_(1))]ubrace)],[(f_(,alpha)v^(alpha)+(1)/(2)f_(,alpha beta)v^(alpha)v^(beta))_(P_(1)),(ubrace(ubrace)_(,alpha,)u^(alpha)+(1)/(2)f_(,alpha beta)u^(alpha)u^(beta))_(P_(0)))-f(P_(0))]],[,-ubrace([f(P_(2))-f(P_(0))]ubrace)]:}quad-ubrace([f(P_(3))-f(P_(2))]ubrace)]\begin{array}{rl} f\left(\mathscr{P}_{4}\right)-f\left(\mathscr{P}_{3}\right)= & \underbrace{\left[f\left(\mathscr{P}_{4}\right)-f\left(\mathscr{P}_{1}\right)\right]} \\ \left(f_{, \alpha} v^{\alpha}+\frac{1}{2} f_{, \alpha \beta} v^{\alpha} v^{\beta}\right)_{\mathscr{P}_{1}} & (\underbrace{}_{, \alpha,} u^{\alpha}+\frac{1}{2} f_{, \alpha \beta} u^{\alpha} u^{\beta})_{\mathscr{P}_{0}})-f\left(\mathscr{P}_{0}\right)] \\ & -\underbrace{\left[f\left(\mathscr{P}_{2}\right)-f\left(\mathscr{P}_{0}\right)\right]} \end{array} \quad-\underbrace{\left[f\left(\mathscr{P}_{3}\right)-f\left(\mathscr{P}_{2}\right)\right]}]f(P4)f(P3)=[f(P4)f(P1)](f,αvα+12f,αβvαvβ)P1(,α,uα+12f,αβuαuβ)P0)f(P0)][f(P2)f(P0)][f(P3)f(P2)]]
Here "cubic errors" are cut down by a factor of 8 , while [ u , v ] f [ u , v ] f [u,v]f[\boldsymbol{u}, \boldsymbol{v}] f[u,v]f is cut down by one of 4 , whenever u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are cut in half.
3. The result
f ( P 4 ) f ( P 3 ) = { [ u , v ] [ f ] } O 0 + "cubic errors" f P 4 f P 3 = { [ u , v ] [ f ] } O 0 +  "cubic errors"  f(P_(4))-f(P_(3))={[u,v][f]}_(O_(0))+" "cubic errors" "f\left(\mathscr{P}_{4}\right)-f\left(\mathscr{P}_{3}\right)=\{[\boldsymbol{u}, \boldsymbol{v}][f]\}_{\mathscr{\mathscr { O }}_{0}}+\text { "cubic errors" }f(P4)f(P3)={[u,v][f]}O0+ "cubic errors" 
says that [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] is a tangent vector at P 0 P 0 P_(0)\mathscr{P}_{0}P0 that describes the separation between the points P 3 P 3 P_(3)\mathscr{P}_{3}P3 and P 4 P 4 P_(4)\mathscr{P}_{4}P4. Its description gets arbitrarily accurate when u u u\boldsymbol{u}u and v v v\boldsymbol{v}v get arbitrarily short. Thus, [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] closes the quadrilateral whose edges are the projections of u u u\boldsymbol{u}u and v v v\boldsymbol{v}v into spacetime.

C. Philosophy of Pictures

  1. Pictures are no substitute for computation. Rather, they are useful for (a) suggesting geometric relationships that were previously unsuspected and that one verifies subsequently by computation; (b) interpreting newly learned geometric results.
  2. This usual noncomputational role of pictures permits one to be sloppy in drawing them. No essential new insight was gained in part B over part A, when one carefully moved the tangent vectors into their respective tangent spaces, and permitted only curves to lie in spacetime. Moreover, the original picture (part A) was clearer because of its greater simplicity.
  3. This motivates one to draw "sloppy" pictures, with tangent vectors lying in spacetime itself-so long as one keeps those tangent vectors short and occasionally checks the scaling of errors when the lengths of the vectors are halved.
Conversely, if one is given a field of basis vectors ("frame field") { e α ( P ) } e α ( P ) {e_(alpha)(P)}\left\{\boldsymbol{e}_{\alpha}(\mathscr{P})\right\}{eα(P)}, but one does not know whether a coordinate system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)} exists in which { e α } = { / x α } e α = / x α {e_(alpha)}={del//delx^(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}=\left\{\partial / \partial x^{\alpha}\right\}{eα}={/xα}, one can find out by a simple test: calculate all ( 4 × 3 ) / 2 = 6 ( 4 × 3 ) / 2 = 6 (4xx3)//2=6(4 \times 3) / 2=6(4×3)/2=6 commutators [ e α , e β ] e α , e β [e_(alpha),e_(beta)]\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right][eα,eβ]; if they all vanish, then there exists such a coordinate system. If not, there doesn't. Stated more briefly, { e α ( P ) } e α ( P ) {e_(alpha)(P)}\left\{\boldsymbol{e}_{\alpha}(\mathscr{P})\right\}{eα(P)} is a coordinate-induced basis if and only if [ e α , e β ] = 0 e α , e β = 0 [e_(alpha),e_(beta)]=0\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]=0[eα,eβ]=0 for all e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα and e β e β e_(beta)\boldsymbol{e}_{\beta}eβ. (See exercise 9.9 for proof; see § 11.5 § 11.5 §11.5\S 11.5§11.5 for an important application.) Coordinate-induced bases are sometimes called "holonomic." In an "anholonomic basis" (noncoordinate basis), one defines the commutation coefficients c μ ν α c μ ν α c_(mu nu)^(alpha)c_{\mu \nu}{ }^{\alpha}cμνα by
(9.22) [ e μ , e ν ] = c μ ν α e α . (9.22) e μ , e ν = c μ ν α e α . {:(9.22)[e_(mu),e_(nu)]=c_(mu nu)^(alpha)e_(alpha).:}\begin{equation*} \left[\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right]=c_{\mu \nu}{ }^{\alpha} \boldsymbol{e}_{\alpha} . \tag{9.22} \end{equation*}(9.22)[eμ,eν]=cμναeα.
They enter into the component formula for the commutator of arbitrary vector fields u u u\boldsymbol{u}u and v v v\boldsymbol{v}v :
(9.23) [ u , v ] = ( u [ v β ] v [ u β ] + u μ v ν c μ ν β ) e β (9.23) [ u , v ] = u v β v u β + u μ v ν c μ ν β e β {:(9.23)[u","v]=(u[v^(beta)]-v[u^(beta)]+u^(mu)v^(nu)c_(mu nu)^(beta))e_(beta):}\begin{equation*} [\boldsymbol{u}, \boldsymbol{v}]=\left(\boldsymbol{u}\left[v^{\beta}\right]-\boldsymbol{v}\left[u^{\beta}\right]+u^{\mu} v^{\nu} c_{\mu \nu}{ }^{\beta}\right) \boldsymbol{e}_{\beta} \tag{9.23} \end{equation*}(9.23)[u,v]=(u[vβ]v[uβ]+uμvνcμνβ)eβ
(see exercise 9.10).
[Warning! In notation for functions and fields, mathematicians and physicists often use the same symbols to mean contradictory things. The physicist may write \ell when considering the length of some critical component in an instrument he is designing, then switch to ( T ) ( T ) ℓ(T)\ell(T)(T) when he begins to analyze its response to temperature changes. Thus \ell is a number, whereas ( T ) ( T ) ℓ(T)\ell(T)(T) is a function. The mathematician, in contrast, will write f f fff for a function that he may be considering as an element in some infinite-dimensional function space. Once the function is supplied with an argument, he then contemplates f ( x ) f ( x ) f(x)f(x)f(x), which is merely a number: the value of f f fff at the point x x xxx. Caught between these antithetical rituals of the physics and mathematics sects, the authors have adopted a clear policy: vacillation. Usually physics-sect statements, like "On a curve P ( λ ) P ( λ ) P(lambda)dots\mathscr{P}(\lambda) \ldotsP(λ)," are used; and the reader can translate them himself into mathematically precise language: "Consider a curve C C C\mathcal{C}C on which a typical point is P = C ( λ ) P = C ( λ ) P=C(lambda)\mathscr{P}=\mathcal{C}(\lambda)P=C(λ); on this curve ..." But on occasion the reader will encounter a pedan-tic-sounding paragraph written in mathematics-sect jargon (Example: Box 23.3). Such paragraphs deal with concepts and relationships so complex that standard physics usage would lead to extreme confusion. They also should prevent the reader from becoming so conditioned to physics usage that he is allergic to the mathematical literature, where great advantages of clarity and economy of thought are achieved by consistent reliance on wholly unambiguous notation.]

Exercise 9.7. PRACTICE WITH COMMUTATORS

Compute the commutator [ e θ ^ , e β ^ ] e θ ^ , e β ^ [e_( hat(theta)),e_( hat(beta))]\left[\boldsymbol{e}_{\hat{\theta}}, \boldsymbol{e}_{\hat{\beta}}\right][eθ^,eβ^] of the vector fields
e θ ^ = 1 r θ , e ϕ ^ = 1 r sin θ ϕ . e θ ^ = 1 r θ , e ϕ ^ = 1 r sin θ ϕ . e_( hat(theta))=(1)/(r)(del)/(del theta),quade_( hat(phi))=(1)/(r sin theta)(del)/(del phi).\boldsymbol{e}_{\hat{\theta}}=\frac{1}{r} \frac{\partial}{\partial \theta}, \quad \boldsymbol{e}_{\hat{\phi}}=\frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} .eθ^=1rθ,eϕ^=1rsinθϕ.
Express your result as a linear combination of e θ ¯ e θ ¯ e_( bar(theta))\boldsymbol{e}_{\bar{\theta}}eθ¯ and e ϕ ˙ e ϕ ˙ e_(phi^(˙))\boldsymbol{e}_{\dot{\phi}}eϕ˙.
Vanishing commutator: a test for coordinate bases
Commutation coefficients defined
Physicists' notation vs. mathematicians' notation

Exercise 9.8. ANGULAR MOMENTUM OPERATORS

In Cartesian coordinates of three-dimensional Euclidean space, one defines three "angularmomentum operators" (vector fields) L j L j L_(j)\boldsymbol{L}_{j}Lj by
L j ϵ j k x k ( / x i ) L j ϵ j k x k / x i L_(j)-=epsilon_(jk)x^(k)(del//delx^(i))\boldsymbol{L}_{j} \equiv \boldsymbol{\epsilon}_{j k} x^{k}\left(\partial / \partial x^{i}\right)Ljϵjkxk(/xi)
Draw a picture of these three vector fields. Calculate their commutators both pictorially and analytically.

Exercise 9.9. COMMUTATORS AND COORDINATE-INDUCED BASES

Let u u u\boldsymbol{u}u and v v v\boldsymbol{v}v be vector fields in spacetime. Show that in some neighborhood of any given point there exists a coordinate system for which
u = / x 1 , v = / x 2 u = / x 1 , v = / x 2 u=del//delx^(1),quad v=del//delx^(2)\boldsymbol{u}=\partial / \partial x^{1}, \quad \boldsymbol{v}=\partial / \partial x^{2}u=/x1,v=/x2
if and only if u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are linearly independent and commute:
[ u , v ] = 0 [ u , v ] = 0 [u,v]=0[\boldsymbol{u}, \boldsymbol{v}]=0[u,v]=0
First make this result plausible from the second figure in Box 9.2 ; then prove it mathematically. Note: this result can be generalized to four arbitrary vector fields e 0 , e 1 , e 2 , e 3 e 0 , e 1 , e 2 , e 3 e_(0),e_(1),e_(2),e_(3)\boldsymbol{e}_{0}, \boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e0,e1,e2,e3. There exists a coordinate system in which e α = / x α e α = / x α e_(alpha)=del//delx^(alpha)\boldsymbol{e}_{\alpha}=\partial / \partial x^{\alpha}eα=/xα if and only if e 0 , e 1 , e 2 , e 3 e 0 , e 1 , e 2 , e 3 e_(0),e_(1),e_(2),e_(3)\boldsymbol{e}_{0}, \boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e0,e1,e2,e3 are linearly independent and [ e μ , e ν ] = 0 e μ , e ν = 0 [e_(mu),e_(nu)]=0\left[\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right]=0[eμ,eν]=0 for all pairs e μ , e ν e μ , e ν e_(mu),e_(nu)\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}eμ,eν.
Exercise 9.10. COMPONENTS OF COMMUTATOR IN NON-COORDINATE BASIS Derive equation (9.23).

Exercise 9.11. LIE DERIVATIVE

The "Lie derivative" of a vector field v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathscr{P})v(P) along a vector field u ( P ) u ( P ) u(P)\boldsymbol{u}(\mathscr{P})u(P) is defined by
(9.24) w u v [ u , v ] . (9.24) w u v [ u , v ] . {:(9.24)w_(u)v-=[u","v].:}\begin{equation*} \mathscr{\mathscr { w }}_{u} \boldsymbol{v} \equiv[\boldsymbol{u}, \boldsymbol{v}] . \tag{9.24} \end{equation*}(9.24)wuv[u,v].
Draw a space-filling family of curves (a "congruence") on a sheet of paper. Draw an arbitrary vector v v v\boldsymbol{v}v at an arbitrary point P 0 P 0 P_(0)\mathscr{P}_{0}P0 on the sheet. Transport that vector along the curve through P 0 P 0 P_(0)\mathscr{P}_{0}P0 by means of the "Lie transport law" u v = 0 u v = 0 ""⧸""_(u)v=0\boldsymbol{\not}_{\boldsymbol{u}} \boldsymbol{v}=0uv=0, where u = d / d t u = d / d t u=d//dt\boldsymbol{u}=d / d tu=d/dt is the tangent to the curve. Draw the resulting vector v v v\boldsymbol{v}v at various points P ( t ) P ( t ) P(t)\mathscr{P}(t)P(t) along the curve.

Exercise 9.12. A CHIP OFF THE OLD BLOCK

(a) Prove the Jacobi identity
(9.25) [ u , [ v , w ] ] + [ v , [ w , u ] ] + [ w , [ u , v ] ] = 0 (9.25) [ u , [ v , w ] ] + [ v , [ w , u ] ] + [ w , [ u , v ] ] = 0 {:(9.25)[u","[v","w]]+[v","[w","u]]+[w","[u","v]]=0:}\begin{equation*} [\boldsymbol{u},[\boldsymbol{v}, \boldsymbol{w}]]+[\boldsymbol{v},[\boldsymbol{w}, \boldsymbol{u}]]+[\boldsymbol{w},[\boldsymbol{u}, \boldsymbol{v}]]=0 \tag{9.25} \end{equation*}(9.25)[u,[v,w]]+[v,[w,u]]+[w,[u,v]]=0
by picking out all terms of the form u v w u v w del_(u)del_(v)del_(w)\partial_{\boldsymbol{u}} \partial_{\boldsymbol{v}} \partial_{\boldsymbol{w}}uvw, showing that they add to zero, and arguing from symmetry that all other terms, e.g., w u v w u v del_(w)del_(u)del_(v)\partial_{\boldsymbol{w}} \partial_{\boldsymbol{u}} \partial_{\boldsymbol{v}}wuv terms, must similarly cancel.
(b) State this identity in index form.
(c) Draw a picture corresponding to this identity (see Box 9.2).

§9.7. MANIFOLDS AND DIFFERENTIAL TOPOLOGY

Spacetime is not the only arena in which the ideas of this chapter can be applied. Points, curves, vectors, 1-forms, and tensors exist in any "differentiable manifold."
Their use to study differentiable manifolds constitutes a branch of mathematics called "differential topology"-hence the title of this chapter.
The mathematician usually begins his development of differential topology by introducing some very primitive concepts, such as sets and topologies of sets, by building a fairly elaborate framework out of them, and by then using that framework to define the concept of a differentiable manifold. But most physicists are satisfied with a more fuzzy, intuitive definition of manifold: roughly speaking, an n n nnn-dimensional differentiable manifold is a set of "points" tied together continuously and differentiably, so that the points in any sufficiently small region can be put into a one-to-one correspondence with an open set of points of R n R n R^(n)R^{n}Rn. [ R n R n R^(n)R^{n}Rn is the number space of n n nnn dimensions, i.e., the space of ordered n n nnn-tuples ( x 1 , x 2 , , x n ) x 1 , x 2 , , x n (x^(1),x^(2),dots,x^(n))\left(x^{1}, x^{2}, \ldots, x^{n}\right)(x1,x2,,xn).] That correspondence furnishes a coordinate system for the neighborhood.
A few examples will convey the concept better than this definition. Elementary examples (Euclidean 3 -spaces, the surface of a sphere) bring to mind too many geometric ideas from richer levels of geometry; so one is forced to contemplate something more complicated. Let R 3 R 3 R^(3)R^{3}R3 be a three-dimensional number space with the usual advanced-calculus ideas of continuity and differentiability. Points ξ ξ xi\boldsymbol{\xi}ξ of R 3 R 3 R^(3)R^{3}R3 are triples, ξ = ( ξ 1 , ξ 2 , ξ 3 ) ξ = ξ 1 , ξ 2 , ξ 3 xi=(xi_(1),xi_(2),xi_(3))\boldsymbol{\xi}=\left(\xi_{1}, \xi_{2}, \xi_{3}\right)ξ=(ξ1,ξ2,ξ3), of real numbers. Let a ray P P P\mathscr{P}P in R 3 R 3 R^(3)R^{3}R3 be any half-line from the origin consisting of all ξ ξ xi\xiξ of the form ξ = λ η ξ = λ η xi=lambda eta\xi=\lambda \boldsymbol{\eta}ξ=λη for some fixed η 0 η 0 eta!=0\boldsymbol{\eta} \neq 0η0 and for all positive real numbers λ > 0 λ > 0 lambda > 0\lambda>0λ>0. (See Figure 9.3.) A good example of a differentiable manifold then is the set S 2 S 2 S^(2)S^{2}S2 of all distinct rays. If f f fff is a real-valued function with a specific value f ( P ) f ( P ) f(P)f(\mathscr{P})f(P) for any ray P P P\mathscr{P}P [so one writes f : S 2 R : P f ( P ) f : S 2 R : P f ( P ) f:S^(2)longrightarrow R:Plongrightarrow f(P)f: S^{2} \longrightarrow R: \mathscr{P} \longrightarrow f(\mathscr{P})f:S2R:Pf(P) ], it should be intuitively (or even demonstrably) clear that we can define what we mean by saying f f fff is continuous or differentiable. In this sense S 2 S 2 S^(2)S^{2}S2 itself is continuous and differentiable. Thus S 2 S 2 S^(2)S^{2}S2 is a manifold, and the rays P P P\mathscr{P}P are the points of S 2 S 2 S^(2)S^{2}S2. There are many other manifolds that differential topology finds indistinguishable from S 2 S 2 S^(2)S^{2}S2. The simplest is the two-dimensional spherical surface ( 2 -sphere), which is the standard representation of S 2 S 2 S^(2)S^{2}S2; it is the set of points ξ ξ xi\boldsymbol{\xi}ξ of R 3 R 3 R^(3)R^{3}R3 satisfying ( ξ 1 ) 2 + ( ξ 2 ) 2 + ( ξ 3 ) 2 = 1 ξ 1 2 + ξ 2 2 + ξ 3 2 = 1 (xi_(1))^(2)+(xi_(2))^(2)+(xi_(3))^(2)=1\left(\xi_{1}\right)^{2}+\left(\xi_{2}\right)^{2}+\left(\xi_{3}\right)^{2}=1(ξ1)2+(ξ2)2+(ξ3)2=1. Clearly a different point P P P\mathscr{P}P of S 2 S 2 S^(2)S^{2}S2 (one ray in R 3 R 3 R^(3)R^{3}R3 ) intersects each point of this standard 2 -sphere surface, and the correspondence is continuous and differentiable in either direction (ray to point; point to ray). The same is true for any ellipsoidal surface in R 3 R 3 R^(3)R^{3}R3 enclosing the origin, and for any other surface enclosing the origin that has
Figure 9.3.
Three different representations of the differentiable manifold S 2 S 2 S^(2)S^{2}S2. The first is the set of all rays emanating from the origin; the second is the sphere they intersect; the third is an oddshaped, closed surface that each ray intersects precisely once.
The manifold S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) (rotation group)
Affine geometry and Riemannian geometry defined
a different ray through each point of itself. They each embody the same global continuity and differentiability concepts, and represent the same abstract differentiable manifold S 2 S 2 S^(2)S^{2}S2, the 2 -sphere. They, and the bundle of rays we started with, all have the same geometric properties at this rudimentary level of geometry. A two-dimensional manifold that has a different geometric structure at this level (a different "differentiable structure") is the torus T 2 T 2 T^(2)T^{2}T2, the surface of a donut. There is no way to imbed this surface smoothly in R 3 R 3 R^(3)R^{3}R3 so that a distinct ray P S 2 P S 2 PinS^(2)\mathscr{P} \in S^{2}PS2 intersects each of its points; there is no invertible and differentiable correspondence between T 2 T 2 T^(2)T^{2}T2 and S 2 S 2 S^(2)S^{2}S2.
Another example of a manifold is the rotation group S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3), whose points P P P\mathscr{P}P are all the 3 × 3 3 × 3 3xx33 \times 33×3 orthogonal matrices of unit determinant, so P = P i j P = P i j P=||P_(ij)||\mathscr{P}=\left\|P_{i j}\right\|P=Pij with P T P = 1 P T P = 1 P^(T)P=1\mathscr{P}^{T} \mathscr{P}=1PTP=1 and det P = 1 det P = 1 detP=1\operatorname{det} \mathscr{P}=1detP=1. This is a three-dimensional space (one often uses the three Eulerangle parameters in computations), where differential ideas (e.g., angular velocity) are employed; hence, it is a manifold. So is the Lorentz group.
The differentiability of a manifold (i.e., the possibility of defining differentiable functions on it) permits one to introduce coordinate systems locally, if not globally, and also curves, tangent spaces, tangent vectors, 1-forms, and tensors, just as is done for spacetime. But the mere fact that a manifold is differentiable does not mean that such concepts as geodesics, parallel transport, curvature, metric, or length exist in it. These are additional layers of structure possessed by some manifolds, but not by all. Roughly speaking, every manifold has smoothness properties and topology, but without additional structure it is shapeless and sizeless.
That branch of mathematics which adds geodesics, parallel transport, and curvature (shape) to a manifold is called affine geometry; that branch which adds a metric is called Riemannian geometry. They will be studied in the next few chapters.

EXERCISES

EXERCISES ON THE ROTATION GROUP

As the exposition of differential geometry becomes more and more sophisticated in the following chapters, the exercises will return time and again to the rotation group as an example of a manifold. Then, in Box 30.1, the results developed in these exercises will be used to analyze the "Mixmaster universe," which is a particularly important cosmological solution to Einstein's field equation.
Before working these exercises, the reader may wish to review the Euler-angle parametrization for rotation matrices, as treated, e.g., on pp. 107-109 of Goldstein (1959).

Exercise 9.13. ROTATION GROUP: GENERATORS

Let K 1 K 1 K_(1)\mathscr{K}_{1}K1 be three 3 × 3 3 × 3 3xx33 \times 33×3 matrices whose components are ( K ) m n = ϵ imn K m n = ϵ imn  (K_(ℓ))_(mn)=epsilon_("imn ")\left(K_{\ell}\right)_{m n}=\epsilon_{\text {imn }}(K)mn=ϵimn .
(a) Display the matrices K 1 , ( K 1 ) 2 , ( K 1 ) 3 K 1 , K 1 2 , K 1 3 K_(1),(K_(1))^(2),(K_(1))^(3)\mathscr{K}_{1},\left(\mathscr{K}_{1}\right)^{2},\left(\mathscr{K}_{1}\right)^{3}K1,(K1)2,(K1)3, and ( K 1 ) 4 K 1 4 (K_(1))^(4)\left(\mathscr{K}_{1}\right)^{4}(K1)4.
(b) Sum the series
(9.26) R x ( θ ) exp ( K 1 θ ) = n = 0 θ n n ! ( K 1 ) n . (9.26) R x ( θ ) exp K 1 θ = n = 0 θ n n ! K 1 n . {:(9.26)R_(x)(theta)-=exp(K_(1)theta)=sum_(n=0)^(oo)(theta^(n))/(n!)(K_(1))^(n).:}\begin{equation*} \mathscr{R}_{x}(\theta) \equiv \exp \left(\mathscr{K}_{1} \theta\right)=\sum_{n=0}^{\infty} \frac{\theta^{n}}{n!}\left(\mathscr{K}_{1}\right)^{n} . \tag{9.26} \end{equation*}(9.26)Rx(θ)exp(K1θ)=n=0θnn!(K1)n.
Show that R x ( θ ) R x ( θ ) R_(x)(theta)\mathscr{R}_{x}(\theta)Rx(θ) is a rotation matrix and that it produces a rotation through an angle θ θ theta\thetaθ about the x x xxx-axis.
(c) Show similarly that R z ( ϕ ) = exp ( K 3 ϕ ) R z ( ϕ ) = exp K 3 ϕ R_(z)(phi)=exp(K_(3)phi)\mathscr{R}_{z}(\phi)=\exp \left(\mathscr{K}_{3} \phi\right)Rz(ϕ)=exp(K3ϕ) and R y ( χ ) = exp ( K 2 χ ) R y ( χ ) = exp K 2 χ R_(y)(chi)=exp(K_(2)chi)\mathscr{R}_{y}(\chi)=\exp \left(\mathscr{K}_{2} \chi\right)Ry(χ)=exp(K2χ) are rotation matrices, and that they produce rotations through angles ϕ ϕ phi\phiϕ and χ χ chi\chiχ about the z z zzz - and y y yyy-axes, respectively.
(d) Explain why P = R z ( ψ ) R x ( θ ) R z ( ϕ ) P = R z ( ψ ) R x ( θ ) R z ( ϕ ) P=R_(z)(psi)R_(x)(theta)R_(z)(phi)\mathscr{P}=\mathscr{R}_{z}(\psi) \mathscr{R}_{x}(\theta) \mathscr{R}_{z}(\phi)P=Rz(ψ)Rx(θ)Rz(ϕ) defines the Euler-angle coordinates, ψ , θ , ϕ ψ , θ , ϕ psi,theta,phi\psi, \theta, \phiψ,θ,ϕ for the generic element P S O ( 3 ) P S O ( 3 ) Pin SO(3)\mathscr{P} \in S O(3)PSO(3) of the rotation group.
(e) Let C C C\mathcal{C}C be the curve P = R z ( t ) P = R z ( t ) P=R_(z)(t)\mathscr{P}=\mathscr{R}_{z}(t)P=Rz(t) through the identity matrix, E ( 0 ) = G S O ( 3 ) E ( 0 ) = G S O ( 3 ) E(0)=Gin SO(3)\mathcal{E}(0)=\mathscr{G} \in S O(3)E(0)=GSO(3). Show that its tangent, ( d C / d t ) ( 0 ) C ˙ ( 0 ) ( d C / d t ) ( 0 ) C ˙ ( 0 ) (dC//dt)(0)-=C^(˙)(0)(d \mathcal{C} / d t)(0) \equiv \dot{\mathcal{C}}(0)(dC/dt)(0)C˙(0) does not vanish by computing C ˙ ( 0 ) f 12 C ˙ ( 0 ) f 12 C^(˙)(0)f_(12)\dot{\mathcal{C}}(0) f_{12}C˙(0)f12, where f 12 f 12 f_(12)f_{12}f12 is the function f 12 ( P ) = P 12 f 12 ( P ) = P 12 f_(12)(P)=P_(12)f_{12}(\mathscr{P})=P_{12}f12(P)=P12, whose value is the 12 matrix element of P P P\mathscr{P}P.
(f) Define a vector field e 3 e 3 e_(3)\boldsymbol{e}_{3}e3 on S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) by letting e 3 ( P ) e 3 ( P ) e_(3)(P)\boldsymbol{e}_{3}(\mathscr{P})e3(P) be the tangent (at t = 0 t = 0 t=0t=0t=0 ) to the curve E ( t ) = R z ( t ) P E ( t ) = R z ( t ) P E(t)=R_(z)(t)P\mathcal{E}(t)=\mathscr{R}_{z}(t) \mathscr{P}E(t)=Rz(t)P through P P P\mathscr{P}P. Show that e 3 ( P ) e 3 ( P ) e_(3)(P)\boldsymbol{e}_{3}(\mathscr{P})e3(P) is nowhere zero. Note: e 3 ( P ) e 3 ( P ) e_(3)(P)\boldsymbol{e}_{3}(\mathscr{P})e3(P) is called the "generator of rotations about the z z zzz-axis," because it points from P P P\mathscr{P}P toward neighboring rotations, R z ( t ) P R z ( t ) P R_(z)(t)P\mathscr{R}_{z}(t) \mathscr{P}Rz(t)P, which differ from P P P\mathscr{P}P by a rotation about the z z zzz-axis.
(g) Show that e 3 = ( / ψ ) θ ϕ e 3 = ( / ψ ) θ ϕ e_(3)=(del//del psi)_(theta phi)\boldsymbol{e}_{3}=(\partial / \partial \psi)_{\theta \phi}e3=(/ψ)θϕ.
(h) Derive the following formulas, valid for t 1 t 1 t≪1t \ll 1t1 :
R x ( t ) R z ( ψ ) x ( θ ) R z ( ϕ ) = z ( ψ t sin ψ cot θ ) R x ( θ + t cos ψ ) R z ( ϕ + t sin ψ / sin θ ) R y ( t ) R z ( ψ ) x ( θ ) R z ( ϕ ) = R z ( ψ + t cos ψ cot θ ) R x ( θ + t sin ψ ) R z ( ϕ t cos ψ / sin θ ) . R x ( t ) R z ( ψ ) x ( θ ) R z ( ϕ ) = z ( ψ t sin ψ cot θ ) R x ( θ + t cos ψ ) R z ( ϕ + t sin ψ / sin θ ) R y ( t ) R z ( ψ ) x ( θ ) R z ( ϕ ) = R z ( ψ + t cos ψ cot θ ) R x ( θ + t sin ψ ) R z ( ϕ t cos ψ / sin θ ) . {:[R_(x)(t)R_(z)(psi)ℜ_(x)(theta)R_(z)(phi)=ℜ_(z)(psi-t sin psi cot theta)R_(x)(theta+t cos psi)R_(z)(phi+t sin psi//sin theta)],[R_(y)(t)R_(z)(psi)ℜ_(x)(theta)R_(z)(phi)=R_(z)(psi+t cos psi cot theta)R_(x)(theta+t sin psi)R_(z)(phi-t cos psi//sin theta).]:}\begin{aligned} & \mathscr{R}_{x}(t) \mathscr{R}_{z}(\psi) \Re_{x}(\theta) \mathscr{R}_{z}(\phi)=\Re_{z}(\psi-t \sin \psi \cot \theta) \mathscr{R}_{x}(\theta+t \cos \psi) \mathscr{R}_{z}(\phi+t \sin \psi / \sin \theta) \\ & \mathscr{R}_{y}(t) \mathscr{R}_{z}(\psi) \Re_{x}(\theta) \mathscr{R}_{z}(\phi)=\mathscr{R}_{z}(\psi+t \cos \psi \cot \theta) \mathscr{R}_{x}(\theta+t \sin \psi) \mathscr{R}_{z}(\phi-t \cos \psi / \sin \theta) . \end{aligned}Rx(t)Rz(ψ)x(θ)Rz(ϕ)=z(ψtsinψcotθ)Rx(θ+tcosψ)Rz(ϕ+tsinψ/sinθ)Ry(t)Rz(ψ)x(θ)Rz(ϕ)=Rz(ψ+tcosψcotθ)Rx(θ+tsinψ)Rz(ϕtcosψ/sinθ).
(i) Define e 1 ( P ) e 1 ( P ) e_(1)(P)\boldsymbol{e}_{1}(\mathscr{P})e1(P) and e 2 ( P ) e 2 ( P ) e_(2)(P)\boldsymbol{e}_{2}(\mathscr{P})e2(P) to be the tangent vectors (at t = 0 t = 0 t=0t=0t=0 ) to the curves E ( t ) = R x ( t ) P E ( t ) = R x ( t ) P E(t)=R_(x)(t)P\mathcal{E}(t)=\mathscr{R}_{x}(t) \mathscr{P}E(t)=Rx(t)P and C ( t ) = R y ( t ) P C ( t ) = R y ( t ) P C(t)=R_(y)(t)P\mathcal{C}(t)=\mathscr{R}_{y}(t) \mathscr{P}C(t)=Ry(t)P, respectively. Show that
e 1 = cos ψ θ sin ψ ( cot θ ψ 1 sin θ ϕ ˙ ) e 2 = sin ψ θ + cos ψ ( cot θ ψ 1 sin θ ϕ ) e 1 = cos ψ θ sin ψ cot θ ψ 1 sin θ ϕ ˙ e 2 = sin ψ θ + cos ψ cot θ ψ 1 sin θ ϕ {:[e_(1)=cos psi(del)/(del theta)-sin psi(cot theta(del)/(del psi)-(1)/(sin theta)(del)/(del(phi^(˙))))],[e_(2)=sin psi(del)/(del theta)+cos psi(cot theta(del)/(del psi)-(1)/(sin theta)(del)/(del phi))]:}\begin{aligned} & \boldsymbol{e}_{1}=\cos \psi \frac{\partial}{\partial \theta}-\sin \psi\left(\cot \theta \frac{\partial}{\partial \psi}-\frac{1}{\sin \theta} \frac{\partial}{\partial \dot{\phi}}\right) \\ & \boldsymbol{e}_{2}=\sin \psi \frac{\partial}{\partial \theta}+\cos \psi\left(\cot \theta \frac{\partial}{\partial \psi}-\frac{1}{\sin \theta} \frac{\partial}{\partial \phi}\right) \end{aligned}e1=cosψθsinψ(cotθψ1sinθϕ˙)e2=sinψθ+cosψ(cotθψ1sinθϕ)
e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 and e 2 e 2 e_(2)\boldsymbol{e}_{2}e2 are the "generators of rotations about the x x xxx - and y y yyy-axes."

Exercise 9.14. ROTATION GROUP: STRUCTURE CONSTANTS

Use the three vector fields constructed in the last exercise,
e 1 = cos ψ θ sin ψ ( cot θ ψ 1 sin θ ϕ ) (9.27) e 2 = sin ψ θ + cos ψ ( cot θ ψ 1 sin θ ϕ ˙ ) e 3 = ψ e 1 = cos ψ θ sin ψ cot θ ψ 1 sin θ ϕ (9.27) e 2 = sin ψ θ + cos ψ cot θ ψ 1 sin θ ϕ ˙ e 3 = ψ {:[e_(1)=cos psi(del)/(del theta)-sin psi(cot theta(del)/(del psi)-(1)/(sin theta)(del)/(del phi))],[(9.27)e_(2)=sin psi(del)/(del theta)+cos psi(cot theta(del)/(del psi)-(1)/(sin theta)(del)/(del(phi^(˙))))],[e_(3)=(del)/(del psi)]:}\begin{align*} & \boldsymbol{e}_{1}=\cos \psi \frac{\partial}{\partial \theta}-\sin \psi\left(\cot \theta \frac{\partial}{\partial \psi}-\frac{1}{\sin \theta} \frac{\partial}{\partial \phi}\right) \\ & \boldsymbol{e}_{2}=\sin \psi \frac{\partial}{\partial \theta}+\cos \psi\left(\cot \theta \frac{\partial}{\partial \psi}-\frac{1}{\sin \theta} \frac{\partial}{\partial \dot{\phi}}\right) \tag{9.27}\\ & \boldsymbol{e}_{3}=\frac{\partial}{\partial \psi} \end{align*}e1=cosψθsinψ(cotθψ1sinθϕ)(9.27)e2=sinψθ+cosψ(cotθψ1sinθϕ˙)e3=ψ
as basis vectors for the manifold of the rotation group. The above equations express this "basis of generators" in terms of the Euler-angle basis. Show that the commutation coefficients for this basis are
(9.28) c α β γ = ϵ α β γ (9.28) c α β γ = ϵ α β γ {:(9.28)c_(alpha beta)^(gamma)=-epsilon_(alpha beta gamma):}\begin{equation*} c_{\alpha \beta}^{\gamma}=-\boldsymbol{\epsilon}_{\alpha \beta \gamma} \tag{9.28} \end{equation*}(9.28)cαβγ=ϵαβγ
independently of location P P P\mathscr{P}P in the rotation group. These coefficients are also called the structure constants of the rotation group.
снартев 10

AFFINE GEOMETRY: GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE

Galilei's Principle of Inertia is sufficient in itself to prove

conclusively that the world is affine in character.
hermann Weyl
This chapter is entirely Track 2. Chapter 9 is necessary preparation for it.
It will be needed as preparation for
(1) Chapters 11-13
(differential geometry;
Newtonian gravity),
(2) the second half of Chapter 14 (calculation of curvature), and
(3) the details, but not the message, of Chapter 15 (Bianchi identities).
Freely falling particles and their clocks

§10.1. GEODESICS AND THE EQUIVALENCE PRINCIPLE

Free fall is the "natural state of motion," so natural, in fact, that the path through spacetime of a freely falling, neutral test body is independent of its structure and composition (the "weak equivalence principle" of Einstein, Eötvös, Dicke; see Box 1.2 and § 38.3 § 38.3 §38.3\S 38.3§38.3 ).
Picture spacetime as filled with free-fall trajectories. Pick an event. Pick a velocity there. They determine a unique trajectory.
Be more precise. Ask for the maximum amount of information tied up in each trajectory. Is it merely the sequence of points along which the test body falls? No; there is more. Each test body can carry a clock with itself (same kind of clock-"good" clock in sense of Figure 1.9-regardless of structure or composition of test body). The clock ticks as the body moves, labeling each event on its trajectory with a number: the time λ λ lambda\lambdaλ the body was there. Result: the free-fall trajectory is not just a sequence of points; it is a parametrized sequence, a "curve" P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ).
But is the parametrization unique? Not entirely. Quite arbitrary are (1) the choice of time origin, P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0); and (2) the units (centimeters, seconds, furlongs, . . .) in which clock time λ λ lambda\lambdaλ is measured. Hence, λ λ lambda\lambdaλ is unique only up to linear transformations
(10.1) λ new = a λ old + b ; (10.1) λ new  = a λ old  + b ; {:(10.1)lambda_("new ")=alambda_("old ")+b;:}\begin{equation*} \lambda_{\text {new }}=a \lambda_{\text {old }}+b ; \tag{10.1} \end{equation*}(10.1)λnew =aλold +b;
Figure 10.1.
A geodesic viewed as a rule for "straight-on parallel transport." Pick an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 and a tangent vector u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ there. Construct the unique geodesic P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) that (1) passes through P 0 : P ( 0 ) = P 0 P 0 : P ( 0 ) = P 0 P_(0):P(0)=P_(0)\mathscr{P}_{0}: \mathscr{P}(0)=\mathscr{P}_{0}P0:P(0)=P0; and (2) has u u u\boldsymbol{u}u as its tangent vector there: ( d O / d λ ) λ = 0 = u ( d O / d λ ) λ = 0 = u (dO//d lambda)_(lambda=0)=u(d \mathscr{\mathscr { O }} / d \lambda)_{\lambda=0}=\boldsymbol{u}(dO/dλ)λ=0=u. This geodesic can be viewed as a rule for picking up u u u\boldsymbol{u}u from P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0) and laying it down again at its tip, P ( 1 ) P ( 1 ) P(1)\mathscr{P}(1)P(1), in as straight a manner as possible,
u λ = 1 = ( d P / d λ ) λ = 1 u λ = 1 = ( d P / d λ ) λ = 1 u_(lambda=1)=(dP//d lambda)_(lambda=1)\boldsymbol{u}_{\lambda=1}=(d \mathscr{P} / d \lambda)_{\lambda=1}uλ=1=(dP/dλ)λ=1
and for then picking it up and laying it down as straight as possible again at L ( 2 ) L ( 2 ) L(2)\mathscr{\mathscr { L }}(2)L(2),
u λ = 2 = ( d P / d λ ) λ = 2 u λ = 2 = ( d P / d λ ) λ = 2 u_(lambda=2)=(dP//d lambda)_(lambda=2)\boldsymbol{u}_{\lambda=2}=(d \mathscr{P} / d \lambda)_{\lambda=2}uλ=2=(dP/dλ)λ=2
etc. This sequence of "straight as possible," "tail-on-tip" transports gives meaning to the idea that ( d P / d λ ) λ = 17 ( d P / d λ ) λ = 17 (dP//d lambda)_(lambda=17)(d \mathscr{P} / d \lambda)_{\lambda=17}(dP/dλ)λ=17 and u = ( d P / d λ ) λ = 0 u = ( d P / d λ ) λ = 0 u=(dP//d lambda)_(lambda=0)\boldsymbol{u}=(d \mathscr{P} / d \lambda)_{\lambda=0}u=(dP/dλ)λ=0 are "the same vector" at different points along the geodesic; or, equivalently, that one has been obtained from the other by "straight-on parallel transport."
b b bbb ("new origin of clock time") is a number independent of location on this specific free-fall trajectory, and a a aaa ("ratio of new units to old") is also.
In the curved spacetime of Einstein (and in that of Cartan-Newton, Chapter 12), these parametrized free-fall trajectories are the straightest of all possible curves. Consequently, one gives these trajectories the same name, "geodesics," that mathematicians use for the straight lines of a curved manifold; and like the mathematicians, one uses the name "affine parameter" for the parameter λ λ lambda\lambdaλ along a free-fall geodesic. Equation (10.1) then says "the affine parameter of a geodesic is unique up to linear transformations."
The affine parameter ("clock time") along a geodesic has nothing to do, à priori, with any metric. It exists even in the absence of metric (e.g., in Cartan-Newtonian spacetime). It gives one a method for comparing the separation between events on a geodesic ( B B B\mathscr{B}B and Q Q Q\mathscr{Q}Q are "twice as far apart" as R R R\mathscr{R}R and Q Q Q\mathscr{Q}Q if [ λ A λ Q ] = 2 [ λ R λ Q ] ) λ A λ Q = 2 λ R λ Q {:[lambda_(A)-lambda_(Q)]=2[lambda_(R)-lambda_(Q)])\left.\left[\lambda_{\mathscr{A}}-\lambda_{\mathscr{Q}}\right]=2\left[\lambda_{\mathscr{R}}-\lambda_{\mathscr{Q}}\right]\right)[λAλQ]=2[λRλQ]). But the affine parameter measures relative separations only along its own geodesic. It has no means of reaching off the geodesic.
The above features of geodesics, and others, are summarized in Figure 10.1 and Box 10.1.

§10.2. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE: PICTORIAL APPROACH

Two test bodies, initially falling through spacetime on parallel, neighboring geodesics, get pushed toward each other or apart by tidal gravitational forces (spacetime curvature). To quantify this statement, one must quantify the concepts of "parallel" and "rate of acceleration away from each other." Begin with parallelism.
Geodesic defined as a free-fall trajectory
Affine parameter defined as clock time along free-fall trajectory

Box 10.1 GEODESICS

Geodesic in brief
Geodesic: in context of gravitation physics
Comparison of vectors at different events by parallel transport
Give point, give tangent vector; get unique, affine-parametrized curve ("geodesic").
World line of a neutral test particle ("Einstein's geometric theory of gravity"; also "Cartan's translation into geometric terms of Newton's theory of gravity"):
(1) "given point": some event on this world line;
(2) "given vector": vector ("displacement per unit increase of parameter") tangent to world line at instant defined by that event;
(3) "unique curve": every neutral test particle with a specified initial position and a specified initial velocity follows the same world line, regardless of its composition and regardless of its mass (small; test mass!; "weak equivalence principle of Einstein-Eötvös-Dicke");
(4) "affine parameter": in Cartan-Newton theory, Newton's "universal time" (which is measured by "good" clocks); in the real physical world, "proper time" (as measured by a "good" clock) along a timelike geodesic;
(5) "parametrized curve": (a) affine parameter unique up to a transformation of the form λ a λ + b λ a λ + b lambda longrightarrow a lambda+b\lambda \longrightarrow a \lambda+bλaλ+b, where a a aaa and b b bbb are constants (no arbitrariness along a given geodesic other than zero of parameter and unit of parameter); or equivalently (b) given any three events a , B a , B a,Ba, \mathscr{B}a,B, C C C\mathcal{C}C on the geodesic, one can find by well-determined physical construction ("clocking") a unique fourth event D D D\mathscr{D}D on the geodesic such that ( λ D λ e ) λ D λ e (lambda_(D)-lambda_(e))\left(\lambda_{\mathscr{D}}-\lambda_{e}\right)(λDλe) is equal to ( λ λ C ) λ λ C (lambda_(刃刃)-lambda_(C))\left(\lambda_{刃 刃}-\lambda_{\mathscr{C}}\right)(λλC); or equivalently (c) [differential version] given a tangent vector with components ( d x α / d λ ) d d x α / d λ d (dx^(alpha)//d lambda)_(d)\left(d x^{\alpha} / d \lambda\right)_{d}(dxα/dλ)d at point a a a\mathscr{a}a, one can find by physical construction (again "clocking") "the same tangent vector" at point C C C\mathcal{C}C with uniquely determined components ( d x α / d λ ) E d x α / d λ E (dx^(alpha)//d lambda)_(E)\left(d x^{\alpha} / d \lambda\right)_{\mathcal{E}}(dxα/dλ)E (vector "equal"; components ordinarily not equal because of twisting and turning of arbitrary base vectors between a a aaa and C C C\mathcal{C}C ).
Consider two neighboring events a a a\mathscr{a}a and B B B\mathscr{B}B connected by a curve P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ). A vector v G v G v_(G)\boldsymbol{v}_{\mathscr{G}}vG lies in the tangent space at A A A\mathscr{A}A, and a vector v B v B v_(B)\boldsymbol{v}_{\mathscr{B}}vB lies in the tangent space at B B B\mathscr{B}B. How can one say whether v d v d v_(d)\boldsymbol{v}_{d}vd and v G v G v_(G)\boldsymbol{v}_{\mathscr{G}}vG are parallel, and how can one compare their lengths? The equivalence principle gives an answer: an observer travels (using rocket power as necessary) through spacetime along the world line P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ). He carries the vector v d v d v_(d)\boldsymbol{v}_{d}vd with himself as he moves, and he uses flat-space Newtonian or Minkowskian standards to keep it always unchanging (flat-space physics is valid locally
according to the equivalence principle!).On reaching event G G G\mathscr{G}G the observer compares his"parallel-transported vector" v d v d v_(d)\boldsymbol{v}_{d}vd with the vector v B v B v_(B)\boldsymbol{v}_{B}vB .If they are identical,then the original vector v G v G v_(G)\boldsymbol{v}_{G}vG was(by definition)parallel to v j v j v_(कj)\boldsymbol{v}_{क j}vj ,and they had the same length. (No metric means no way to quantify length;nevertheless,parallel transport gives a way to compare length!)
The equivalence principle entered this discussion in a perhaps unfamiliar way, applied to an observer who may be accelerated,rather than to one who is freely falling.But one cannot evade a basic principle by merely confronting it with an intricate application.(Ingenious perpetual-motion machines are as impossible as simpleminded ones!)The equivalence principle states that no local measurement that is insensitive to gravitational tidal forces can detect any difference whatsoever between flat and curved spacetime.The spaceship navigator has an inertial guidance system(accelerometers,gyroscopes,computers)capable of preserving an inertial reference frame in flat spacetime;and in flat spacetime it can compute the attitude and velocity of any object in the spaceship relative to a given inertial frame.The purchaser may specify whether he wants a guidance computer programmed with the laws of zero-gravity Newtonian mechanics,or with those of special-relativity physics.Use this same guidance system-including the same computer program-in curved spacetime.A vector is being parallel transported if the guidance system's computer says it is not changing.
Will the result of transport in this way be independent of the curve used to link C C C\mathscr{C}C and G G G\mathscr{G}G ?Clearly yes,in gravity-free spacetime,since this is a principal performance criterion that the purchaser of an inertial guidance system can demand of the manufacturer.But in a curved spacetime,the answer is"NO!"If v d v d v_(d)\boldsymbol{v}_{d}vd agrees with v F v F v_(F)\boldsymbol{v}_{\mathscr{F}}vF after parallel transport along one curve,it need not agree with v v v_(刃刃)\boldsymbol{v}_{刃 刃}v after parallel transport along another.Spacetime curvature produces discrepancies.But one is not ready to study and quantify those discrepancies(Chapter 11),until one has developed the mathematical formalism of parallel transport,which,in turn,cannot be done until one has made precise the"flat-space standards for keeping the vector v d v d v_(d)\boldsymbol{v}_{d}vd always unchanging"as it is transported along a curve.
The flat-space standards are made precise in Box 10.2.They lead to(1)a"Schild's ladder"construction for performing parallel transport;(2)the concept"covariant derivative," u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv ,of a vector field v v v\boldsymbol{v}v along a curve with tangent u u u\boldsymbol{u}u ;(3)the"equation of motion" u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 for a geodesic,which states that"a geodesic parallel transports its own tangent vector along itself;"and(4)a link between the tangent spaces at adjacent events(Figure 10.2).

§10.3.PARALLEL TRANSPORT AND COVARIANT DERIVATIVE:ABSTRACT APPROACH

From the"Schild's ladder"construction of Box 10.2,one learns the following properties of spacetime's covariant derivative:
Result of parallel transport depends on route
Schild's ladder for performing parallel transport;its consequences
Parallel transport defined using inertial guidance systems and equivalence principle

Box 10.2 FROM GEODESICS TO PARALLEL TRANSPORT TO COVARIANT DIFFERENTIATION TO GEODESICS TO ...

"Parallel transport" as defined by geodesics


A. Transport any sufficiently short stretch of a curve a X a X aXa \mathscr{X}aX (i.e., any tangent vector) parallel to itself along curve Q G Q G QG\mathscr{Q} \mathscr{G}QG to point B B B\mathscr{B}B as follows:
  1. Take some point π π pi\mathscr{\pi}π along A A A\mathscr{A}A close to a a aaa. Take geodesic X π X π Xpi\mathscr{X} \mathscr{\pi}Xπ through X X X\mathscr{X}X and π π pi\mathscr{\pi}π. Take any affine parametrization λ λ lambda\lambdaλ of X A X A XA\mathscr{X} \mathscr{A}XA and define a unique point R R R\mathscr{R}R by the condition λ O = 1 2 ( λ X + λ K ) λ O = 1 2 λ X + λ K lambda_(O)=(1)/(2)(lambda_(X)+lambda_(K))\lambda_{\mathscr{O}}=\frac{1}{2}\left(\lambda_{\mathscr{X}}+\lambda_{\mathscr{K}}\right)λO=12(λX+λK) ("equal stretches of time in X C X C XC\mathscr{X} \mathscr{\mathscr { C }}XC and R R R\mathscr{\mathscr { R }}R ").
  2. Take geodesic that starts at a a a\mathfrak{a}a and passes through τ τ tau\mathscr{\tau}τ, and extend it by an equal parameter increment to point P P P\mathscr{P}P.
  3. Curve N P N P NP\mathscr{N P}NP gives vector A X A X AX\mathscr{A X}AX as propagated parallel to itself from a a aaa to π π pi\mathscr{\pi}π (for sufficiently short A X A X AX\mathscr{A X}AX and A R ) A R ) AR)\mathscr{A R})AR). This construction certainly yields parallel transport in flat spacetime (Newtonian or Einsteinian). Moreover, it is local (vectors A X , C R A X , C R AX,CR\mathfrak{A X}, \mathbb{C} \mathscr{R}AX,CR, etc., very short). Therefore, it must work even in curved spacetime. (It embodies the equivalence principle.)
  4. Repeat process over and over, and eventually end up with C X C X CX\mathscr{C X}CX propagated parallel to itself from a a aaa to B B B\mathscr{B}B. Call this construction "Schild's Ladder," from Schild's (1970) similar construction. [See also Ehlers, Pirani, and Schild (1972).] Note that curve A G B A G B AGB\mathscr{A G B}AGB need not be a geodesic. There is no requirement that N Q Q N Q Q NQQ\mathscr{N Q} \mathcal{Q}NQQ be the straight-on continuation of A A A A AA\mathscr{A} \mathscr{A}AA similar to the geodesic requirement in the "cross-brace" that P P P P PP\mathscr{\mathscr { P } P}PP be the straight-on continuation of A r A r Ar\mathfrak{A r}Ar.

    "Schild's Ladder"
    "Covariant differentiation" as defined by parallel transport along a curve with tangent vector u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ. The answer, d v / d λ u v d v / d λ u v dv//d lambda-=grad_(u)v-=d \boldsymbol{v} / d \lambda \equiv \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v} \equivdv/dλuv "rate of change of v v v\boldsymbol{v}v with respect to λ λ lambda\lambdaλ " -=\equiv "covariant derivative of v v v\boldsymbol{v}v along u u u\boldsymbol{u}u, " is constructed by the following obvious procedure: (1) Take v v v\boldsymbol{v}v at λ = λ 0 + ε λ = λ 0 + ε lambda=lambda_(0)+epsi\lambda=\lambda_{0}+\varepsilonλ=λ0+ε. (2) Parallel transport it back to λ = λ 0 λ = λ 0 lambda=lambda_(0)\lambda=\lambda_{0}λ=λ0. (3) Calculate how much it differs from v v v\boldsymbol{v}v there. (4) Divide by ε ε epsi\varepsilonε (and take limit as ε 0 ε 0 epsi longrightarrow0\varepsilon \longrightarrow 0ε0 ):
u v = Lim ε 0 { [ v ( λ 0 + ε ) ] parallel transported to λ 0 v ( λ 0 ) ε } . u v = Lim ε 0 v λ 0 + ε parallel transported to  λ 0 v λ 0 ε . grad_(u)v=Lim_(epsi rarr0){([v(lambda_(0)+epsi)]_("parallel transported to "lambda_(0))-v(lambda_(0)))/(epsi)}.\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}=\operatorname{Lim}_{\varepsilon \rightarrow 0}\left\{\frac{\left[\boldsymbol{v}\left(\lambda_{0}+\varepsilon\right)\right]_{\text {parallel transported to } \lambda_{0}}-\boldsymbol{v}\left(\lambda_{0}\right)}{\varepsilon}\right\} .uv=Limε0{[v(λ0+ε)]parallel transported to λ0v(λ0)ε}.
If u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ is short compared to scale of inhomogeneities in the vector field v v v\boldsymbol{v}v, then u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv can be read directly off drawing I, or, equally well, off drawing II.
Box 10.2 (continued)

"Symmetry" of covariant differentiation

Chain rule for covariant differentiation
Additivity for covariant differentiation
C. Take two vector fields. Combine into one the two diagrams for u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv and v u v u grad_(v)u\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}vu. Thereby discover that u v v u u v v u grad_(u)v-grad_(v)u\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}uvvu is the vector by which the v u v u v u v u v-u-v-u\boldsymbol{v} \boldsymbol{-} \boldsymbol{u}-\boldsymbol{v}-\boldsymbol{u}vuvu quadrilateral fails to close-i.e. (see Box 9.2), it is the commutator [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] : u v v u = [ u , v ] u v v u = [ u , v ] grad_(u)v-grad_(v)u=[u,v]\nabla_{u} v-\nabla_{v} u=[u, v]uvvu=[u,v].
Terminology: grad\mathbf{\nabla} is said to be a "symmetric" or "torsion-free" covariant derivative when u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv v u = [ u , v ] v u = [ u , v ] -grad_(v)u=[u,v]-\nabla_{\boldsymbol{v}} \boldsymbol{u}=[\boldsymbol{u}, \boldsymbol{v}]vu=[u,v]. Other types of covariant derivatives, as studied by mathematicians, have no relevance for any gravitation theory based on the equivalence principle.
D. The "take-the-difference" and "take-the-limit" process used to define u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv guarantees that it obeys the usual rule for differentiating products:

(for proof, see exercise 10.2.)
E. In the real physical world, be it Newtonian or relativistic, parallel transport of a triangle cannot break its legs apart: (1) A , B , C A , B , C A,B,C\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}A,B,C initially such that A + B = C A + B = C A+B=C\boldsymbol{A}+\boldsymbol{B}=\boldsymbol{C}A+B=C; (2) A , B , C A , B , C A,B,C\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}A,B,C each parallel transported with himself by freely falling (inertial) observer; (3) then A + B = C A + B = C A+B=C\boldsymbol{A}+\boldsymbol{B}=\boldsymbol{C}A+B=C always. Any other result would violate the equivalence principle!
  1. Consequence of this (as seen by following through definition of covariant derivative, and by noting that any vector u u u\boldsymbol{u}u can be regarded as the tangent vector to a freely falling world line):
u ( v + w ) = u v + u w u ( v + w ) = u v + u w grad_(u)(v+w)=grad_(u)v+grad_(u)w\nabla_{u}(v+w)=\nabla_{u} v+\nabla_{u} \boldsymbol{w}u(v+w)=uv+uw
for any vector u u u\boldsymbol{u}u and vector fields v v v\boldsymbol{v}v and w w w\boldsymbol{w}w.
2. Consequence of this, combined with symmetry of covariant derivative, and with additivity of the "closer of quadrilaterals" [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] :
u + n v = u v + n v u + n v = u v + n v grad_(u+n)v=grad_(u)v+grad_(n)v\nabla_{u+n} v=\nabla_{u} v+\nabla_{n} vu+nv=uv+nv
(See exercise 10.1.) This can be inferred, alternatively, from the equivalence principle: in a local inertial frame, as in special relativity or Newtonian theory, the change in v v v\boldsymbol{v}v along u + n u + n u+n\boldsymbol{u}+\boldsymbol{n}u+n should equal the sum of the changes along u u u\boldsymbol{u}u and along n n n\boldsymbol{n}n.
3. Consequence of above: choose n n n\boldsymbol{n}n to be a multiple of u u u\boldsymbol{u}u; thereby conclude
a u v = a u v . a u v = a u v . grad_(au)v=agrad_(u)v.\boldsymbol{\nabla}_{a u} \boldsymbol{v}=a \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v} .auv=auv.
F. The "Schild's ladder" construction process for parallel transport (beginning of this box), applied to the tangent vector of a geodesic (exercise 10.6) guarantees: a geodesic parallel transports its own tangent vector along itself. Translated into covariant-derivative language:
( u = d / d λ is a tangent vector to a curve, and u u = 0 ) ( the curve is a geodesic ) . u = d / d λ  is a tangent   vector to a curve, and  u u = 0 (  the curve is   a geodesic  ) . ([u=d//d lambda" is a tangent "],[" vector to a curve, and "],[grad_(u)u=0])=>((" the curve is ")/(" a geodesic ")).\left(\begin{array}{l} \boldsymbol{u}=d / d \lambda \text { is a tangent } \\ \text { vector to a curve, and } \\ \boldsymbol{\nabla}_{u} \boldsymbol{u}=0 \end{array}\right) \Rightarrow\binom{\text { the curve is }}{\text { a geodesic }} .(u=d/dλ is a tangent  vector to a curve, and uu=0)( the curve is  a geodesic ).
Thus closes the circle: geodesic to parallel transport to covariant derivative to geodesic.
Geodesics as defined by parallel transport or covariant differentiation
Covariant derivative: basic properties
Symmetry: u v v u = [ u , v ] for any vector fields u and v ; Chain rule: u ( f v ) = f u v + v u f for any function f , vector field v , and vector u ; Additivity: u ( v + w ) = u v + u w for any vector fields v and w , and vector u ; a u + b n v = a u v + b n v for any vector field v , vectors or vector fields u and n , and numbers or functions a and b .  Symmetry:  u v v u = [ u , v ]  for any vector fields  u  and  v  Chain rule:  u ( f v ) = f u v + v u f  for any function  f  vector field  v , and vector  u  Additivity:  u ( v + w ) = u v + u w  for any vector   fields  v  and  w , and vector  u a u + b n v = a u v + b n v  for any vector   field  v , vectors or vector fields  u  and  n  and numbers or functions  a  and  b {:[" Symmetry: "quadgrad_(u)v-grad_(v)u=[u","v]" for any vector fields "u" and "v"; "],[" Chain rule: "quadgrad_(u)(fv)=fgrad_(u)v+vdel_(u)f" for any function "f", "],[" vector field "v", and vector "u"; "],[" Additivity: "],[grad_(u)(v+w)=grad_(u)v+grad_(u)w" for any vector "],[" fields "v" and "w", and vector "u"; "],[grad_(au+bn)v=agrad_(u)v+bgrad_(n)v" for any vector "],[" field "v", vectors or vector fields "u" and "n", "],[" and numbers or functions "a" and "b". "]:}\begin{aligned} & \text { Symmetry: } \quad \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}=[\boldsymbol{u}, \boldsymbol{v}] \text { for any vector fields } \boldsymbol{u} \text { and } \boldsymbol{v} \text {; } \\ & \text { Chain rule: } \quad \boldsymbol{\nabla}_{\boldsymbol{u}}(f \boldsymbol{v})=f \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}+\boldsymbol{v} \partial_{\boldsymbol{u}} f \text { for any function } f \text {, } \\ & \text { vector field } \boldsymbol{v} \text {, and vector } \boldsymbol{u} \text {; } \\ & \text { Additivity: } \\ & \boldsymbol{\nabla}_{u}(\boldsymbol{v}+\boldsymbol{w})=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}+\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w} \text { for any vector } \\ & \text { fields } \boldsymbol{v} \text { and } \boldsymbol{w} \text {, and vector } \boldsymbol{u} \text {; } \\ & \boldsymbol{\nabla}_{a \boldsymbol{u}+b \boldsymbol{n}} \boldsymbol{v}=a \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}+b \boldsymbol{\nabla}_{\boldsymbol{n}} \boldsymbol{v} \text { for any vector } \\ & \text { field } \boldsymbol{v} \text {, vectors or vector fields } \boldsymbol{u} \text { and } \boldsymbol{n} \text {, } \\ & \text { and numbers or functions } a \text { and } b \text {. } \end{aligned} Symmetry: uvvu=[u,v] for any vector fields u and v Chain rule: u(fv)=fuv+vuf for any function f vector field v, and vector u Additivity: u(v+w)=uv+uw for any vector  fields v and w, and vector uau+bnv=auv+bnv for any vector  field v, vectors or vector fields u and n and numbers or functions a and b
Figure 10.2.
The link between the tangent spaces at neighboring points, made possible by a parallel-transport law. Choose basis vectors e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 and e 2 e 2 e_(2)\boldsymbol{e}_{2}e2 at the event a a aaa. Parallel transport them to a neighboring event 9 β 9 β 9beta9 \boldsymbol{\beta}9β. (Schild's ladder for transport of e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 is shown in the figure.) Then any other vector v v v\boldsymbol{v}v that is parallel transported from a a a\mathscr{a}a to B B B\mathscr{B}B will have the same components at the two events (parallel transport cannot break the legs of a triangle; see Box 10.2):
v = v 1 e 1 + v 2 e 2 at a v = v 1 e 1 + v 2 e 2 at B . v = v 1 e 1 + v 2 e 2  at  a v = v 1 e 1 + v 2 e 2  at  B v=v^(1)e_(1)+v^(2)e_(2)" at "a=>v=v^(1)e_(1)+v^(2)e_(2)" at "B". "uarr\boldsymbol{v}=v^{1} \boldsymbol{e}_{1}+v^{2} \boldsymbol{e}_{2} \text { at } a \Rightarrow \underset{\uparrow}{\boldsymbol{v}=v^{1} \boldsymbol{e}_{1}+v^{2} \boldsymbol{e}_{2} \text { at } \mathscr{B} \text {. }}v=v1e1+v2e2 at av=v1e1+v2e2 at B
Thus, parallel transport provides a unique and complete link between the tangent space at a a aaa and the tangent space at B B B\mathscr{B}B. It identifies a unique vector at B B B\mathscr{B}B with each vector at A A A\mathscr{A}A in a way that preserves all algebraic relations. Similarly (see § 10.3 § 10.3 §10.3\S 10.3§10.3 ), it identifies a unique 1 -form at B B B\mathscr{B}B with each 1 -form at A A A\mathscr{A}A, and a unique tensor at 93 with each tensor at d d ddd, preserving all algebraic relations such as σ , v = 19.9 σ , v = 19.9 (:sigma,v:)=19.9\langle\boldsymbol{\sigma}, \boldsymbol{v}\rangle=19.9σ,v=19.9 and S ( σ , v , w ) = 371 S ( σ , v , w ) = 371 S(sigma,v,w)=371\boldsymbol{S}(\boldsymbol{\sigma}, \boldsymbol{v}, \boldsymbol{w})=371S(σ,v,w)=371.
Actually, all this is true only in the limit when A A A\mathscr{A}A and B B B\mathscr{B}B are arbitrarily close to each other. When a a aaa and B B B\mathscr{B}B are close but not arbitrarily close, the result of parallel transport is slightly different for different paths; so the link between the tangent spaces is slightly nonunique. But the differences decrease by a factor of 4 each time the affine-parameter distance between a a aaa and B B B\mathscr{B}B is cut in half; see Chapter 11 .
Any "rule" grad\boldsymbol{\nabla}, for producing new vector fields from old, that satisfies these four conditions, is called by differential geometers a "symmetric covariant derivative." Such a rule is not inherent in the more primitive concepts (Chapter 9) of curves, vectors, tensors, etc. In the arena of a spacetime laboratory, there are as many ways of defining a covariant derivative rule grad\boldsymbol{\nabla} as there are of rearranging sources of the gravitational field. Different free-fall trajectories (geodesics) result from different distributions of masses.
Given the geodesics of spacetime, or of any other manifold, one can construct a unique corresponding covariant derivative by the Schild's ladder procedure of Box 10.2. Given any covariant derivative, one can discuss parallel transport via the equation
(10.3) d v / d λ u v = 0 the vector field v is parallel transported along the vector u = d / d λ (10.3) d v / d λ u v = 0  the vector field  v  is parallel transported   along the vector  u = d / d λ {:[(10.3)dv//d lambda-=grad_(u)v=0Longleftrightarrow" the vector field "v" is parallel transported "],[" along the vector "u=d//d lambda]:}\begin{align*} d \boldsymbol{v} / d \lambda \equiv \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}=0 \Longleftrightarrow & \text { the vector field } \boldsymbol{v} \text { is parallel transported } \tag{10.3}\\ & \text { along the vector } \boldsymbol{u}=d / d \lambda \end{align*}(10.3)dv/dλuv=0 the vector field v is parallel transported  along the vector u=d/dλ
and one can test whether any curve is a geodesic via
u u = 0 the curve P ( λ ) with tangent vector u = d / d λ (10.4) parallel transports its own tangent vector u P ( λ ) is a geodesic. u u = 0  the curve  P ( λ )  with tangent vector  u = d / d λ (10.4)  parallel transports its own tangent vector  u P ( λ )  is a geodesic.  {:[grad_(u)u=0 Longleftrightarrow" the curve "P(lambda)" with tangent vector "u=d//d lambda],[(10.4)" parallel transports its own tangent vector "u],[ LongleftrightarrowP(lambda)" is a geodesic. "]:}\begin{align*} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0 & \Longleftrightarrow \text { the curve } \mathscr{P}(\lambda) \text { with tangent vector } \boldsymbol{u}=d / d \lambda \\ & \text { parallel transports its own tangent vector } \boldsymbol{u} \tag{10.4}\\ & \Longleftrightarrow \mathscr{P}(\lambda) \text { is a geodesic. } \end{align*}uu=0 the curve P(λ) with tangent vector u=d/dλ(10.4) parallel transports its own tangent vector uP(λ) is a geodesic. 
Thus a knowledge of all geodesics is completely equivalent to a knowledge of the covariant derivative.
The covariant derivative grad\boldsymbol{\nabla} generalizes to curved spacetime the flat-space gradient grad\boldsymbol{\nabla}. Like its flat-space cousin, it can be viewed as a machine for producing a number σ , u v σ , u v (:sigma,grad_(u)v:)\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangleσ,uv out of a 1-form σ σ sigma\boldsymbol{\sigma}σ, a vector u u u\boldsymbol{u}u, and a vector field v v v\boldsymbol{v}v. This machine viewpoint is explored in Box 10.3. Note there an important fact: despite its machine nature, grad\boldsymbol{\nabla} is not a tensor; it is a nontensorial geometric object.
In curved as in flat spacetime, grad\boldsymbol{\nabla} can be applied not only to vector fields, but also to functions, 1 -form fields, and tensor fields. Its action on functions is defined in the obvious manner:
(10.5) f d f ; u f u f u [ f ] d f , u . (10.5) f d f ; u f u f u [ f ] d f , u . {:(10.5)grad f-=df;quadgrad_(u)f-=del_(u)f-=u[f]-=(:df","u:).:}\begin{equation*} \boldsymbol{\nabla} f \equiv \boldsymbol{d} f ; \quad \boldsymbol{\nabla}_{\boldsymbol{u}} f \equiv \partial_{\boldsymbol{u}} f \equiv \boldsymbol{u}[f] \equiv\langle\boldsymbol{d} f, \boldsymbol{u}\rangle . \tag{10.5} \end{equation*}(10.5)fdf;ufufu[f]df,u.
Its action on 1-form fields and tensor fields is defined by the curved-space generalization of equation (3.39): S S grad S\boldsymbol{\nabla} \boldsymbol{S}S is a linear machine for calculating the change in output of S S S\boldsymbol{S}S, from point to point, when "constant" (i.e., parallel transported) vectors are inserted into its slots. Example: the gradient of a ( 0 1 ) ( 0 1 ) ((0)/(1))\binom{0}{1}(01) tensor, i.e., of a 1-form field σ σ sigma\boldsymbol{\sigma}σ. Pick an event P 0 P 0 P_(0)\mathscr{P}_{0}P0; pick two vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v in the tangent space at P 0 P 0 P_(0)\mathscr{P}_{0}P0; construct from v v v\boldsymbol{v}v a "constant" vector field v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathscr{P})v(P) by parallel transport along the direction of u , u v = 0 u , u v = 0 u,grad_(u)v=0\boldsymbol{u}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}=0u,uv=0. Then σ σ grad sigma\boldsymbol{\nabla} \boldsymbol{\sigma}σ is a ( 0 2 ) ( 0 2 ) ((0)/(2))\binom{0}{2}(02) tensor, and u σ u σ grad_(u)sigma\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\sigma}uσ is a ( 0 1 ) ( 0 1 ) ((0)/(1))\binom{0}{1}(01) tensor defined at P 0 P 0 P_(0)\mathscr{P}_{0}P0 by
(10.6) σ ( v , u ) u σ , v u ( σ , v ) d d λ σ , v (10.6) σ ( v , u ) u σ , v u ( σ , v ) d d λ σ , v {:(10.6)grad sigma(v","u)-=(:grad_(u)sigma,v:)-=grad_(u)((:sigma","v:))-=(d)/(d lambda)(:sigma","v:):}\begin{equation*} \boldsymbol{\nabla} \sigma(\boldsymbol{v}, \boldsymbol{u}) \equiv\left\langle\boldsymbol{\nabla}_{u} \boldsymbol{\sigma}, \boldsymbol{v}\right\rangle \equiv \boldsymbol{\nabla}_{u}(\langle\boldsymbol{\sigma}, \boldsymbol{v}\rangle) \equiv \frac{d}{d \lambda}\langle\boldsymbol{\sigma}, \boldsymbol{v}\rangle \tag{10.6} \end{equation*}(10.6)σ(v,u)uσ,vu(σ,v)ddλσ,v
where u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ. This defines σ σ grad sigma\boldsymbol{\nabla} \boldsymbol{\sigma}σ and u σ u σ grad_(u)sigma\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\sigma}uσ, because it states their output for any
Knowledge of all geodesics is equivalent to knowledge of covariant derivative
Covariant derivative generalizes flat-space gradient
Action of covariant derivative on functions, 1 -forms, and tensors

Box 10.3 COVARIANT DERIVATIVE VIEWED AS A MACHINE; CONNECTION COEFFICIENTS AS ITS COMPONENTS

A. The Machine View

  1. The covariant derivative operator grad\boldsymbol{\nabla}, like most other geometric objects, can be regarded as a machine with slots. There is one such machine at each event P 0 P 0 P_(0)\mathscr{P}_{0}P0 in spacetime. In brief, the machine interpretation of grad\boldsymbol{\nabla} at P 0 P 0 P_(0)\mathscr{\mathscr { P }}_{0}P0 says

    [Note: this slot notation for grad\boldsymbol{\nabla} serves no useful purpose except to emphasize the "machine"-nature of grad\boldsymbol{\nabla}. This box is the only place it will be used.]
  2. Geometrically, the output of the machine, σ , u v σ , u v (:sigma,grad_(u)v:)\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangleσ,uv, is obtained as follows:
    (a) Calculate the rate of change of v , u v v , u v v,grad_(u)v\boldsymbol{v}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}v,uv, along the vector u u u\boldsymbol{u}u; when u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are infinitesimally small, the calculation can be represented pictorially:

    (b) Count how many surfaces of the 1 -form σ σ sigma\boldsymbol{\sigma}σ are pierced by the vector u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv (piercing occurs in tangent space at P 0 P 0 P_(0)\mathscr{P}_{0}P0 )
σ , u v = 2.8 σ , u v = 2.8 (:sigma,grad_(u)v:)=-2.8\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangle=-2.8σ,uv=2.8
This number is the output of the machine grad\boldsymbol{\nabla}, when σ , v ( P ) σ , v ( P ) sigma,v(P)\boldsymbol{\sigma}, \boldsymbol{v}(\mathscr{P})σ,v(P) and u u u\boldsymbol{u}u are inserted into its slots.
3. Another, equivalent, statement of covariant derivative as a machine. Leave first slot empty (no mention of any 1-form σ σ sigma\boldsymbol{\sigma}σ ); get a new vector field from original vector field v v v\boldsymbol{v}v :
empty ( , , v ( P ) , u ) u v empty  ( , , v ( P ) , u ) u v ubrace(gradubrace)_("empty ")(ubrace(,ubrace),v(P),u)-=grad_(u)v\underbrace{\boldsymbol{\nabla}}_{\text {empty }}(\underbrace{,}, \boldsymbol{v}(\mathscr{P}), \boldsymbol{u}) \equiv \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}empty (,,v(P),u)uv
= "covariant derivative of vector field v along vector u . " =  "covariant derivative of vector field  v  along vector  u . " =" "covariant derivative of vector field "v" along vector "u."=\text { "covariant derivative of vector field } \boldsymbol{v} \text { along vector } \boldsymbol{u} . "= "covariant derivative of vector field v along vector u."
  1. A third machine operation. Leave first and third slots empty (no mention of any 1 -form σ σ sigma\boldsymbol{\sigma}σ; no mention of any vector u u u\boldsymbol{u}u along which to differentiate); get a ( ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11) tensor field from original vector field v v v\boldsymbol{v}v :
( empty , v ( P ) , empty ) v = "covariant derivative" or "gradient" of vector field v . ( empty  , v ( P ) , empty  ) v =  "covariant derivative" or "gradient" of vector field  v . {:[grad(ubrace(dotsubrace)_("empty ")","v(P)","ubrace(dotsubrace)_("empty "))-=gradv],[=" "covariant derivative" or "gradient" of vector field "v.]:}\begin{aligned} \boldsymbol{\nabla}(\underbrace{\ldots}_{\text {empty }}, \boldsymbol{v}(\mathscr{P}), \underbrace{\ldots}_{\text {empty }}) & \equiv \mathbf{\nabla} \boldsymbol{v} \\ & =\text { "covariant derivative" or "gradient" of vector field } \boldsymbol{v} . \end{aligned}(empty ,v(P),empty )v= "covariant derivative" or "gradient" of vector field v.
This tensor field, v v grad v\boldsymbol{\nabla} \boldsymbol{v}v, is the curved-space generalization of the flat-space v v grad v\boldsymbol{\nabla} \boldsymbol{v}v studied in §3.5. It has two slots (the two left empty in its definition). Its output for given input is
v ( , u ) = u v empty v ( σ , u ) = σ , u v . v ( , u ) = u v empty  v ( σ , u ) = σ , u v . {:[grad v(ubrace(dots,u)=grad_(u)vubrace)_("empty ")],[grad v(sigma","u)=(:sigma,grad_(u)v:).]:}\begin{gathered} \boldsymbol{\nabla} \boldsymbol{v}(\underbrace{\ldots, \boldsymbol{u})=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}}_{\text {empty }} \\ \boldsymbol{\nabla} \boldsymbol{v}(\boldsymbol{\sigma}, \boldsymbol{u})=\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{u} \boldsymbol{v}\right\rangle . \end{gathered}v(,u)=uvempty v(σ,u)=σ,uv.
  1. Summary of the quantities defined above:
    (a) grad\boldsymbol{\nabla} is a covariant derivative operator; to get a number from it, insert σ , v ( P ) σ , v ( P ) sigma,v(P)\boldsymbol{\sigma}, \boldsymbol{v}(\mathscr{P})σ,v(P), and u u u\boldsymbol{u}u; the result is σ , u v σ , u v (:sigma,grad_(u)v:)\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangleσ,uv.
    (b) v v grad v\boldsymbol{\nabla} \boldsymbol{v}v is the gradient of v v v\boldsymbol{v}v; to get a number from it, insert σ σ sigma\boldsymbol{\sigma}σ and u u u\boldsymbol{u}u; the result is σ , u v σ , u v (:sigma,grad_(u)v:)\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangleσ,uv [same as in (a)].
    (c) u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv is the covariant derivative of v v v\boldsymbol{v}v along u u u\boldsymbol{u}u; to get a number from it, insert σ σ sigma\boldsymbol{\sigma}σ; the result is σ , u v σ , u v (:sigma,grad_(u)v:)\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangleσ,uv [same as in (a) and (b)].

B. How grad\boldsymbol{\nabla} Differs from a Tensor

The machine grad\boldsymbol{\nabla} differs from a tensor in two ways. (1) The middle slot of grad\boldsymbol{\nabla} will not accept a vector; it demands a vector field-the vector field that is to be differentiated. (2) grad\boldsymbol{\nabla} is not a linear machine (whereas a tensor must be linear!):
Box 10.3 (continued)

C. The "Connection Coefficients" as Components of V V V\mathbf{V}V

Given a tensor S S S\boldsymbol{S}S of rank ( 1 2 ) 1 2 ((1)/(2))\left(\frac{1}{2}\right)(12), a basis of tangent vectors { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} at the event P 0 P 0 P_(0)\mathscr{P}_{0}P0 where S S S\boldsymbol{S}S resides, and the dual basis of 1 -forms { ω α } ω α {omega^(alpha)}\left\{\boldsymbol{\omega}^{\alpha}\right\}{ωα}, one defines the components of S S S\boldsymbol{S}S by
S β γ α S ( ω α , e β , e γ ) S β γ α S ω α , e β , e γ S_(beta gamma)^(alpha)-=S(omega^(alpha),e_(beta),e_(gamma))S_{\beta \gamma}^{\alpha} \equiv \boldsymbol{S}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}\right)SβγαS(ωα,eβ,eγ)
One defines the components of grad\boldsymbol{\nabla} similarly, except that for grad\boldsymbol{\nabla} one needs not only a basis { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} at the event P 0 P 0 P_(0)\mathscr{P}_{0}P0, but also a basis { e α ( P ) } e α ( P ) {e_(alpha)(P)}\left\{\boldsymbol{e}_{\alpha}(\mathscr{P})\right\}{eα(P)} at each event P P P\mathscr{P}P in its neighborhood:
Γ α β γ components of = ( ω α , e β ( P ) , e γ ) ω α , e γ e β ( " α -component of change in basis vector e β , when in evaluating e β one moves from tail to tip of e γ ) . Γ α β γ  components of  = ω α , e β ( P ) , e γ ω α , e γ e β (  "  α -component of change in basis vector  e β ,  when   in evaluating  e β  one moves from tail to tip of  e γ ) . {:[Gamma^(alpha)_(beta gamma)-=" components of "grad=grad(omega^(alpha),e_(beta)(P),e_(gamma))],[-=(:omega^(alpha),grad_(e_(gamma))e_(beta):)],[≃((" " "alpha"-component of change in basis vector "e_(beta)," when ")/(" in evaluating "e_(beta)" one moves from tail to tip of "e_(gamma))).]:}\begin{aligned} \Gamma^{\alpha}{ }_{\beta \gamma} & \equiv \text { components of } \boldsymbol{\nabla}=\boldsymbol{\nabla}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}(\mathscr{P}), \boldsymbol{e}_{\gamma}\right) \\ & \equiv\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{\nabla}_{\boldsymbol{e}_{\gamma}} \boldsymbol{e}_{\beta}\right\rangle \\ & \simeq\binom{\text { " } \alpha \text {-component of change in basis vector } \boldsymbol{e}_{\beta}, \text { when }}{\text { in evaluating } \boldsymbol{e}_{\beta} \text { one moves from tail to tip of } \boldsymbol{e}_{\gamma}} . \end{aligned}Γαβγ components of =(ωα,eβ(P),eγ)ωα,eγeβ( " α-component of change in basis vector eβ, when  in evaluating eβ one moves from tail to tip of eγ).
These components of grad\boldsymbol{\nabla} are called the "connection coefficients" of the basis { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα}. They are the "coordinate representation" of the covariant derivative operator grad\boldsymbol{\nabla}.
The covariant derivative operator grad\nabla and the connection coefficients Γ α μ ν Γ α μ ν Gamma^(alpha)_(mu nu)\Gamma^{\alpha}{ }_{\mu \nu}Γαμν provide different mathematical representations of the same geometric animal? Preposterous! The one animal runs from place to place and barks, or at least bites (takes difference, for example, between vector fields at one place and at a nearby place). The other animal, endowed with forty faces (see exercise 10.9) sits quietly at one spot. It would be difficult for two animals to look more different. Yet they do the same jobs in any world compatible with the equivalence principle: (1) they summarize the properties of all geodesics that go through the point in question; and, so doing, (2) they provide a physical means (parallel transport) to compare the values of vector fields and tensor fields at two neighboring events.
given input vectors v v v\boldsymbol{v}v and u u u\boldsymbol{u}u. If v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathscr{P})v(P) is not constrained to be "constant" along u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ, then ( d / d λ ) σ , v ( d / d λ ) σ , v (d//d lambda)(:sigma,v:)(d / d \lambda)\langle\boldsymbol{\sigma}, \boldsymbol{v}\rangle(d/dλ)σ,v has contributions from both the change in v v v\boldsymbol{v}v and the change in σ σ sigma\sigmaσ :
(10.7) d d λ σ , v u σ , v = u σ , v + σ , u v (10.7) d d λ σ , v u σ , v = u σ , v + σ , u v {:(10.7)(d)/(d lambda)(:sigma","v:)-=grad_(u)(:sigma","v:)=(:grad_(u)sigma,v:)+(:sigma,grad_(u)v:):}\begin{equation*} \frac{d}{d \lambda}\langle\boldsymbol{\sigma}, \boldsymbol{v}\rangle \equiv \boldsymbol{\nabla}_{u}\langle\boldsymbol{\sigma}, \boldsymbol{v}\rangle=\left\langle\boldsymbol{\nabla}_{u} \boldsymbol{\sigma}, \boldsymbol{v}\right\rangle+\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangle \tag{10.7} \end{equation*}(10.7)ddλσ,vuσ,v=uσ,v+σ,uv
(see exercise 10.3).
Similarly, if S S S\boldsymbol{S}S is a ( 1 2 ) 1 2 ((1)/(2))\left(\frac{1}{2}\right)(12) tensor field, then its gradient S S grad S\boldsymbol{\nabla} \boldsymbol{S}S is a ( 1 3 ) 1 3 ((1)/(3))\left(\frac{1}{3}\right)(13) tensor field defined as follows. Pick an event P 0 P 0 P_(0)\mathscr{\mathscr { P }}_{0}P0; pick three vectors u , v , w u , v , w u,v,w\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}u,v,w, and a 1 -form σ σ sigma\boldsymbol{\sigma}σ in the tangent space at Φ 0 ; Φ 0 ; Phi_(0);\mathscr{\Phi}_{0} ;Φ0; turn v , w v , w v,w\boldsymbol{v}, \boldsymbol{w}v,w, and σ σ sigma\boldsymbol{\sigma}σ into "constant" vector fields and a "constant" 1-form field near P 0 P 0 P_(0)\mathscr{P}_{0}P0 by means of parallel transport ( u v = u w = u σ = 0 u v = u w = u σ = 0 (grad_(u)v=grad_(u)w=grad_(u)sigma=0:}\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\sigma}=0\right.(uv=uw=uσ=0 at P 0 ) P 0 {:P_(0))\left.\mathscr{P}_{0}\right)P0); then define
(10.8) S ( σ , v , w , u ) ( u S ) ( σ , v , w ) u [ S ( σ , v , w ) ] = u [ S ( σ , v , w ) ] (10.8) S ( σ , v , w , u ) u S ( σ , v , w ) u [ S ( σ , v , w ) ] = u [ S ( σ , v , w ) ] {:[(10.8)grad S(sigma","v","w","u)-=(grad_(u)S)(sigma","v","w)-=grad_(u)[S(sigma","v","w)]],[=del_(u)[S(sigma","v","w)]]:}\begin{align*} \boldsymbol{\nabla} \boldsymbol{S}(\sigma, \boldsymbol{v}, \boldsymbol{w}, \boldsymbol{u}) & \equiv\left(\boldsymbol{\nabla}_{u} \boldsymbol{S}\right)(\boldsymbol{\sigma}, \boldsymbol{v}, \boldsymbol{w}) \equiv \boldsymbol{\nabla}_{u}[\boldsymbol{S}(\sigma, \boldsymbol{v}, \boldsymbol{w})] \tag{10.8}\\ & =\partial_{u}[\boldsymbol{S}(\sigma, \boldsymbol{v}, \boldsymbol{w})] \end{align*}(10.8)S(σ,v,w,u)(uS)(σ,v,w)u[S(σ,v,w)]=u[S(σ,v,w)]

Exercise 10.1. ADDITIVITY OF COVARIANT DIFFERENTIATION

Show that the commutator ("closer of quadrilaterals") is additive:
[ u , v + w ] = [ u , v ] + [ u , w ] ; [ u + n , v ] = [ u , v ] + [ n , v ] . [ u , v + w ] = [ u , v ] + [ u , w ] ; [ u + n , v ] = [ u , v ] + [ n , v ] . [u,v+w]=[u,v]+[u,w];quad[u+n,v]=[u,v]+[n,v].[\boldsymbol{u}, \boldsymbol{v}+\boldsymbol{w}]=[\boldsymbol{u}, \boldsymbol{v}]+[\boldsymbol{u}, \boldsymbol{w}] ; \quad[\boldsymbol{u}+\boldsymbol{n}, \boldsymbol{v}]=[\boldsymbol{u}, \boldsymbol{v}]+[\boldsymbol{n}, \boldsymbol{v}] .[u,v+w]=[u,v]+[u,w];[u+n,v]=[u,v]+[n,v].
Use this result, the additivity condition u ( v + w ) = u v + u w u ( v + w ) = u v + u w grad_(u)(v+w)=grad_(u)v+grad_(u)w\boldsymbol{\nabla}_{\boldsymbol{u}}(\boldsymbol{v}+\boldsymbol{w})=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}+\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}u(v+w)=uv+uw, and symmetry of the covariant derivative, u v v u = [ u , v ] u v v u = [ u , v ] grad_(u)v-grad_(v)u=[u,v]\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}=[\boldsymbol{u}, \boldsymbol{v}]uvvu=[u,v], to prove that
u + n v = u v + n v u + n v = u v + n v grad_(u+n)v=grad_(u)v+grad_(n)v\nabla_{u+n} v=\nabla_{u} v+\nabla_{n} vu+nv=uv+nv

Exercise 10.2. CHAIN RULE FOR COVARIANT DIFFERENTIATION

Use pictures, and the "take-the-difference-and-take-the-limit" definition of u v u v grad_(u)v\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}uv (Box 10.2) to show that
(10.9) u ( f v ) = f u v + v u [ f ] . (10.9) u ( f v ) = f u v + v u [ f ] . {:(10.9)grad_(u)(fv)=fgrad_(u)v+vdel_(u)[f].:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{u}}(f \boldsymbol{v})=f \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}+\boldsymbol{v} \partial_{\boldsymbol{u}}[f] . \tag{10.9} \end{equation*}(10.9)u(fv)=fuv+vu[f].

Exercise 10.3. ANOTHER CHAIN RULE

Derive equation (10.7), using the "take-the-difference-and-take-the-limit" definitions of derivatives. Hint: Before taking the differences, parallel transport σ [ P ( λ ) ] σ [ P ( λ ) ] sigma[P(lambda)]\sigma[\mathscr{P}(\lambda)]σ[P(λ)] and v [ P ( λ ) ] v [ P ( λ ) ] v[P(lambda)]\boldsymbol{v}[\mathscr{P}(\lambda)]v[P(λ)] back from P ( λ ) P ( λ ) P(lambda)\mathscr{P}(\lambda)P(λ) to P ( 0 ) P ( 0 ) P(0)\mathscr{P}(0)P(0).

Exercise 10.4. STILL ANOTHER CHAIN RULE

Show that, as in flat spacetime, so also in curved spacetime,
(10.10) u ( v v ) = ( u v ) w + v ( u v ) (10.10) u ( v v ) = u v w + v u v {:(10.10)grad_(u)(v ox v)=(grad_(u)v)ox w+v ox(grad_(u)v):}\begin{equation*} \boldsymbol{\nabla}_{u}(v \otimes \boldsymbol{v})=\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right) \otimes \boldsymbol{w}+\boldsymbol{v} \otimes\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right) \tag{10.10} \end{equation*}(10.10)u(vv)=(uv)w+v(uv)
Write down the more familiar component version of this equation in flat spacetime.
Solution to first part of exercise: Choose 1 -forms σ σ sigma\sigmaσ and ρ ρ rho\rhoρ at the event P 0 P 0 P_(0)\mathscr{P}_{0}P0 in question, and extend them along the vector u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ by parallel transport, u ρ = u σ = 0 u ρ = u σ = 0 grad_(u)rho=grad_(u)sigma=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\rho}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\sigma}=0uρ=uσ=0. Then
[ u ( v w ) ] ( ρ , σ ) = d d λ [ ( v w ) ( ρ , σ ) ] (def of u on a tensor) = d d λ [ ρ , v σ , w ] (def of tensor product " ") = d ρ , v d λ σ , w + ρ , v d σ , w d λ (chain rule for derivatives) = ρ , u v σ , w + ρ , v σ , u w = [ ( u v ) w ] ( ρ , σ ) + [ v ( u w ) ] ( ρ , σ ) (def of tensor product " "). u ( v w ) ( ρ , σ ) = d d λ [ ( v w ) ( ρ , σ ) ]  (def of  u  on a tensor)  = d d λ [ ρ , v σ , w ]  (def of tensor product "   ")  = d ρ , v d λ σ , w + ρ , v d σ , w d λ  (chain rule for derivatives)  = ρ , u v σ , w + ρ , v σ , u w = u v w ( ρ , σ ) + v u w ( ρ , σ )  (def of tensor product "   ").  {:[[grad_(u)(v ox w)](rho","sigma),=(d)/(d lambda)[(v ox w)(rho","sigma)]," (def of "grad_(u)" on a tensor) "],[,=(d)/(d lambda)[(:rho","v:)(:sigma","w:)]," (def of tensor product " "ox" ") "],[,=(d(:rho,v:))/(d lambda)(:sigma","w:)+(:rho","v:)(d(:sigma,w:))/(d lambda)," (chain rule for derivatives) "],[,=(:rho,grad_(u)v:)(:sigma","w:)+(:rho","v:)(:sigma,grad_(u)w:)],[,=[(grad_(u)v)ox w](rho","sigma)+[v ox(grad_(u)w)](rho","sigma)],[" (def of tensor product " "ox" "). "]:}\begin{array}{rlr} {\left[\boldsymbol{\nabla}_{u}(\boldsymbol{v} \otimes \boldsymbol{w})\right](\boldsymbol{\rho}, \boldsymbol{\sigma})} & =\frac{d}{d \lambda}[(\boldsymbol{v} \otimes \boldsymbol{w})(\boldsymbol{\rho}, \boldsymbol{\sigma})] & \text { (def of } \boldsymbol{\nabla}_{u} \text { on a tensor) } \\ & =\frac{d}{d \lambda}[\langle\boldsymbol{\rho}, \boldsymbol{v}\rangle\langle\boldsymbol{\sigma}, \boldsymbol{w}\rangle] & \text { (def of tensor product " } \otimes \text { ") } \\ & =\frac{d\langle\boldsymbol{\rho}, \boldsymbol{v}\rangle}{d \lambda}\langle\boldsymbol{\sigma}, \boldsymbol{w}\rangle+\langle\boldsymbol{\rho}, \boldsymbol{v}\rangle \frac{d\langle\boldsymbol{\sigma}, \boldsymbol{w}\rangle}{d \lambda} & \text { (chain rule for derivatives) } \\ & =\left\langle\boldsymbol{\rho}, \boldsymbol{\nabla}_{u} \boldsymbol{v}\right\rangle\langle\boldsymbol{\sigma}, \boldsymbol{w}\rangle+\langle\boldsymbol{\rho}, \boldsymbol{v}\rangle\left\langle\boldsymbol{\sigma}, \boldsymbol{\nabla}_{u} \boldsymbol{w}\right\rangle \\ & =\left[\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right) \otimes \boldsymbol{w}\right](\boldsymbol{\rho}, \boldsymbol{\sigma})+\left[\boldsymbol{v} \otimes\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}\right)\right](\boldsymbol{\rho}, \boldsymbol{\sigma}) \\ \text { (def of tensor product " } \otimes \text { "). } \end{array}[u(vw)](ρ,σ)=ddλ[(vw)(ρ,σ)] (def of u on a tensor) =ddλ[ρ,vσ,w] (def of tensor product "  ") =dρ,vdλσ,w+ρ,vdσ,wdλ (chain rule for derivatives) =ρ,uvσ,w+ρ,vσ,uw=[(uv)w](ρ,σ)+[v(uw)](ρ,σ) (def of tensor product "  "). 

Exercise 10.5. ONE MORE CHAIN RULE

Show, using techniques similar to those in exercise 10.4, that
(10.11) u ( σ ρ v ) = ( u σ ) ρ v + σ ( u ρ ) v + σ ρ ( u v ) (10.11) u ( σ ρ v ) = u σ ρ v + σ u ρ v + σ ρ u v {:(10.11)grad_(u)(sigma ox rho ox v)=(grad_(u)sigma)ox rho ox v+sigma ox(grad_(u)rho)ox v+sigma ox rho ox(grad_(u)v):}\begin{equation*} \nabla_{u}(\sigma \otimes \rho \otimes v)=\left(\nabla_{u} \sigma\right) \otimes \rho \otimes v+\sigma \otimes\left(\nabla_{u} \rho\right) \otimes v+\sigma \otimes \rho \otimes\left(\boldsymbol{\nabla}_{u} \boldsymbol{v}\right) \tag{10.11} \end{equation*}(10.11)u(σρv)=(uσ)ρv+σ(uρ)v+σρ(uv)

Exercise 10.6. GEODESIC EQUATION

Use the "Schild's ladder" construction process for parallel transport (beginning of Box 10.2) to show that a geodesic parallel transports its own tangent vector along itself (end of Box 10.2).

§10.4. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE: COMPONENT APPROACH

The pictorial approach motivates the mathematics; the abstract approach makes the pictorial ideas precise; but usually one must use the component approach in order to actually do complex calculations.
To work with components, one needs a set of basis vectors { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} and the dual set of basis 1 -forms { ω α } ω α {omega^(alpha)}\left\{\boldsymbol{\omega}^{\alpha}\right\}{ωα}. In flat spacetime a single such basis suffices; all events can use the same Lorentz basis. Not so in curved spacetime! There each event has its own tangent space, and each tangent space requires a basis of its own. As one travels from event to event, comparing their bases via parallel transport, one sees the bases twist and turn. They must do so. In no other way can they accommodate themselves to the curvature of spacetime. Bases at points P 0 P 0 P_(0)\mathscr{P}_{0}P0 and P 1 P 1 P_(1)\mathscr{P}_{1}P1, which are the same when compared by parallel transport along one curve, must differ when compared along another curve (see "Curvature"; Chapter 11).
To quantify the twisting and turning of a "field" of basis vectors { e α ( P ) } e α ( P ) {e_(alpha)(P)}\left\{\boldsymbol{e}_{\alpha}(\mathscr{P})\right\}{eα(P)} and forms { ω α ( P ) } ω α ( P ) {omega^(alpha)(P)}\left\{\boldsymbol{\omega}^{\alpha}(\mathscr{P})\right\}{ωα(P)}, use the covariant derivative. Examine the changes in vector fields along a basis vector e β e β e_(beta)\boldsymbol{e}_{\beta}eβ, abbreviating
(10.12) e β β ( def of β ) ; (10.12) e β β  def of  β ; {:(10.12)grad_(e_(beta))-=grad_(beta)quad(" def of "grad_(beta));:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{e}_{\beta}} \equiv \boldsymbol{\nabla}_{\beta} \quad\left(\text { def of } \boldsymbol{\nabla}_{\beta}\right) ; \tag{10.12} \end{equation*}(10.12)eββ( def of β);
and especially examine the rate of change of some basis vector: β e α β e α grad_(beta)e_(alpha)\boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\alpha}βeα. This rate of change is itself a vector, so it can be expanded in terms of the basis:
(10.13) β e α = e μ Γ μ μ α β ( def of Γ α β μ ) ; (10.13) β e α = e μ Γ μ μ α β def  of  Γ α β μ ; {:(10.13)grad_(beta)e_(alpha)=e_(mu)Gamma^(mu)ubrace(muubrace)_(alpha beta)quad(def" of "Gamma_(alpha beta)^(mu));:}\begin{equation*} \boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\alpha}=\boldsymbol{e}_{\mu} \Gamma^{\mu} \underbrace{\mu}_{\alpha \beta} \quad\left(\operatorname{def} \text { of } \Gamma_{\alpha \beta}^{\mu}\right) ; \tag{10.13} \end{equation*}(10.13)βeα=eμΓμμαβ(def of Γαβμ);
note reversal of order of α and β !  note reversal of order of  α  and  β  !  " note reversal of order of "alpha" and "beta" ! "\text { note reversal of order of } \alpha \text { and } \beta \text { ! } note reversal of order of α and β ! 
and the resultant "connection coefficients" Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ can be calculated by projection on the basis 1 -forms:
(10.14) ω μ , β e α = Γ α β μ . (10.14) ω μ , β e α = Γ α β μ . {:(10.14)(:omega^(mu),grad_(beta)e_(alpha):)=Gamma_(alpha beta)^(mu).:}\begin{equation*} \left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\alpha}\right\rangle=\Gamma_{\alpha \beta}^{\mu} . \tag{10.14} \end{equation*}(10.14)ωμ,βeα=Γαβμ.
(See exercise 10.7; also Box 10.3.) Because the basis 1 -forms are "locked into" the basis vectors ( ω v , e α = δ ν α ) ω v , e α = δ ν α ((:omega^(v),e_(alpha):)=delta^(nu)_(alpha))\left(\left\langle\boldsymbol{\omega}^{v}, \boldsymbol{e}_{\alpha}\right\rangle=\delta^{\nu}{ }_{\alpha}\right)(ωv,eα=δνα), these same connection coefficients Γ ν α β Γ ν α β Gamma^(nu)_(alpha beta)\Gamma^{\nu}{ }_{\alpha \beta}Γναβ tell how the 1 -form basis changes from point to point:
(10.15) β ω ν = Γ v α β ω α , (10.16) β ω v , e α = Γ v α β . (10.15) β ω ν = Γ v α β ω α , (10.16) β ω v , e α = Γ v α β . {:[(10.15)grad_(beta)omega^(nu)=-Gamma^(v)_(alpha beta)omega^(alpha)","],[(10.16)(:grad_(beta)omega^(v),e_(alpha):)=-Gamma^(v)_(alpha beta).]:}\begin{align*} \boldsymbol{\nabla}_{\beta} \boldsymbol{\omega}^{\nu} & =-\Gamma^{v}{ }_{\alpha \beta} \boldsymbol{\omega}^{\alpha}, \tag{10.15}\\ \left\langle\boldsymbol{\nabla}_{\beta} \boldsymbol{\omega}^{v}, \boldsymbol{e}_{\alpha}\right\rangle & =-\Gamma^{v}{ }_{\alpha \beta} . \tag{10.16} \end{align*}(10.15)βων=Γvαβωα,(10.16)βωv,eα=Γvαβ.
(See exercise 10.8.)
The connection coefficients do even more. They allow one to calculate the components of the gradient of an arbitrary tensor S S S\boldsymbol{S}S. In a Lorentz frame of flat spacetime, the components of S S grad S\boldsymbol{\nabla} \boldsymbol{S}S are obtained by letting the basis vectors e α = P / x α = e α = P / x α = e_(alpha)=delP//delx^(alpha)=\boldsymbol{e}_{\alpha}=\partial \mathscr{P} / \partial x^{\alpha}=eα=P/xα= / x α / x α del//delx^(alpha)\partial / \partial x^{\alpha}/xα act on the components of S S S\boldsymbol{S}S. Thus for a ( 1 2 ) 1 2 ((1)/(2))\left(\frac{1}{2}\right)(12) tensor field S S S\boldsymbol{S}S one finds that
S has components S α β γ , δ = x δ [ S α β γ ] . S  has components  S α β γ , δ = x δ S α β γ . grad S" has components "S^(alpha)_(beta gamma,delta)=(del)/(delx^(delta))[S^(alpha)_(beta gamma)].\boldsymbol{\nabla} \boldsymbol{S} \text { has components } S^{\alpha}{ }_{\beta \gamma, \delta}=\frac{\partial}{\partial x^{\delta}}\left[S^{\alpha}{ }_{\beta \gamma}\right] .S has components Sαβγ,δ=xδ[Sαβγ].
Not so in curved spacetime, or even in a non-Lorentz basis in flat spacetime. There the basis vectors turn, twist, expand, and contract, so even if S S S\boldsymbol{S}S were constant ( S = 0 S = 0 grad S=0\boldsymbol{\nabla} \boldsymbol{S}=0S=0 ), its components on the twisting basis vectors would vary. The connection coefficients, properly applied, will compensate for this twisting and turning. As one learns in exercise 10.10 , the components of S S grad S\boldsymbol{\nabla} \boldsymbol{S}S, called S α β γ ; δ S α β γ ; δ S^(alpha)_(beta gamma;delta)S^{\alpha}{ }_{\beta \gamma ; \delta}Sαβγ;δ so that
(10.17) S = S α β γ ; δ e α ω β ω γ ω δ , (10.17) S = S α β γ ; δ e α ω β ω γ ω δ , {:(10.17)grad S=S^(alpha)_(beta gamma;delta)e_(alpha)oxomega^(beta)oxomega^(gamma)oxomega^(delta)",":}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{S}=S^{\alpha}{ }_{\beta \gamma ; \delta} \boldsymbol{e}_{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma} \otimes \boldsymbol{\omega}^{\delta}, \tag{10.17} \end{equation*}(10.17)S=Sαβγ;δeαωβωγωδ,
can be calculated from those of S S S\boldsymbol{S}S by the usual flat-space method, plus a correction applied to each index (i.e., to each basis vector):
Here
(10.19) S α β γ , δ e δ [ S α β γ ] e s S α β γ . (10.19) S α β γ , δ e δ S α β γ e s S α β γ . {:(10.19)S^(alpha)_(beta gamma,delta)-=e_(delta)[S^(alpha)_(beta gamma)]-=del_(e_(s))S^(alpha)_(beta gamma).:}\begin{equation*} S^{\alpha}{ }_{\beta \gamma, \delta} \equiv \boldsymbol{e}_{\delta}\left[S^{\alpha}{ }_{\beta \gamma}\right] \equiv \partial_{\boldsymbol{e}_{s}} S^{\alpha}{ }_{\beta \gamma} . \tag{10.19} \end{equation*}(10.19)Sαβγ,δeδ[Sαβγ]esSαβγ.
Components of the gradient of a tensor field
Connection coefficients defined
Equation (10.18) looks complicated; but it is really very simple, once the pattern has been grasped.
Just as one uses special notation, S α β γ ; δ S α β γ ; δ S^(alpha)_(beta gamma;delta)S^{\alpha}{ }_{\beta \gamma ; \delta}Sαβγ;δ, for the components of S S grad S\boldsymbol{\nabla} \boldsymbol{S}S, so one
Components of the covariant derivative of a tensor field
introduces special notation, D S α β γ / d λ D S α β γ / d λ DS^(alpha)_(beta gamma)//d lambdaD S^{\alpha}{ }_{\beta \gamma} / d \lambdaDSαβγ/dλ, for components of the covariant derivative u S u S grad_(u)S\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{S}uS along u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ :
(10.20) u s = ( D S α β γ / d λ ) e α ω β ω γ ; D S α β γ d λ = S α β γ ; δ u δ = ( S α β γ , δ + correction terms ) u δ . (10.20) u s = D S α β γ / d λ e α ω β ω γ ; D S α β γ d λ = S α β γ ; δ u δ = S α β γ , δ +  correction terms  u δ . {:[(10.20)grad_(u)s=(DS^(alpha)_(beta gamma)//d lambda)e_(alpha)oxomega^(beta)oxomega^(gamma);],[(DS^(alpha)_(beta gamma))/(d lambda)=S^(alpha)_(beta gamma;delta)u^(delta)=(S^(alpha)_(beta gamma,delta)+" correction terms ")u^(delta).]:}\begin{gather*} \boldsymbol{\nabla}_{u} \boldsymbol{s}=\left(D S^{\alpha}{ }_{\beta \gamma} / d \lambda\right) \boldsymbol{e}_{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma} ; \tag{10.20}\\ \frac{D S^{\alpha}{ }_{\beta \gamma}}{d \lambda}=S^{\alpha}{ }_{\beta \gamma ; \delta} u^{\delta}=\left(S^{\alpha}{ }_{\beta \gamma, \delta}+\text { correction terms }\right) u^{\delta} . \end{gather*}(10.20)us=(DSαβγ/dλ)eαωβωγ;DSαβγdλ=Sαβγ;δuδ=(Sαβγ,δ+ correction terms )uδ.
Since for any f f fff
f , δ u δ = u f = d f / d λ f , δ u δ = u f = d f / d λ f_(,delta)u^(delta)=del_(u)f=df//d lambdaf_{, \delta} u^{\delta}=\partial_{\boldsymbol{u}} f=d f / d \lambdaf,δuδ=uf=df/dλ
this reduces to
(10.21) D S α β γ d λ = d S α β γ d λ + S μ β γ Γ α μ δ u δ S α μ γ Γ μ β δ u δ S α β μ Γ μ γ δ u δ . (10.21) D S α β γ d λ = d S α β γ d λ + S μ β γ Γ α μ δ u δ S α μ γ Γ μ β δ u δ S α β μ Γ μ γ δ u δ . {:(10.21)(DS^(alpha)_(beta gamma))/(d lambda)=(dS^(alpha)_(beta gamma))/(d lambda)+S^(mu)_(beta gamma)Gamma^(alpha)_(mu delta)u^(delta)-S^(alpha)_(mu gamma)Gamma^(mu)_(beta delta)u^(delta)-S^(alpha)_(beta mu)Gamma^(mu)_(gamma delta)u^(delta).:}\begin{equation*} \frac{D S^{\alpha}{ }_{\beta \gamma}}{d \lambda}=\frac{d S^{\alpha}{ }_{\beta \gamma}}{d \lambda}+S^{\mu}{ }_{\beta \gamma} \Gamma^{\alpha}{ }_{\mu \delta} u^{\delta}-S^{\alpha}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta} u^{\delta}-S^{\alpha}{ }_{\beta \mu} \Gamma^{\mu}{ }_{\gamma \delta} u^{\delta} . \tag{10.21} \end{equation*}(10.21)DSαβγdλ=dSαβγdλ+SμβγΓαμδuδSαμγΓμβδuδSαβμΓμγδuδ.
The power of the component approach shows up clearly when one discusses chain rules for covariant derivatives. The multitude of abstract-approach chain rules (equations 10.2 b , 10.7 , 10.10 , 10.11 10.2 b , 10.7 , 10.10 , 10.11 10.2b,10.7,10.10,10.1110.2 \mathrm{~b}, 10.7,10.10,10.1110.2 b,10.7,10.10,10.11 ) all boil down into a single rule for components: The gradient operation ";" obeys the standard partial-differentiation chain rule of ordinary calculus. Example:
(10.22a) ( f v α ) ; μ = f ; μ u α + f v α ; μ [ = f , μ because f has no indices to correct ] (10.22a) f v α ; μ = f ; μ u α + f v α ; μ [ = f , μ  because  f  has no indices to correct  ] {:[(10.22a)(fv^(alpha))_(;mu)=f_(;mu)u^(alpha)+fv^(alpha)_(;mu)],[ ubrace([ubrace)=f_(,mu)" because "f" has no indices to correct "]]:}\begin{align*} \left(f v^{\alpha}\right)_{; \mu}= & f_{; \mu} u^{\alpha}+f v^{\alpha}{ }_{; \mu} \tag{10.22a}\\ & \underbrace{[ }=f_{, \mu} \text { because } f \text { has no indices to correct }] \end{align*}(10.22a)(fvα);μ=f;μuα+fvα;μ[=f,μ because f has no indices to correct ]
(contract this with u μ u μ u^(mu)u^{\mu}uμ to get chain rule 10.2 b ). Another example:
(10.22b) ( σ α v α ) ; μ = σ α ; μ v α + σ α v α ; μ 4 = ( σ α v α ) , μ because σ α v α has no free indices to correct ] (10.22b) σ α v α ; μ = σ α ; μ v α + σ α v α ; μ 4 = σ α v α , μ  because  σ α v α  has no free indices to correct  {:[(10.22b)(sigma_(alpha)v^(alpha))_(;mu)=sigma_(alpha;mu)v^(alpha)+sigma_(alpha)v^(alpha)_(;mu)],[{:uarr^(4)=(sigma_(alpha)v^(alpha))_(,mu)" because "sigma_(alpha)v^(alpha)" has no free indices to correct "]]:}\begin{align*} & \left(\sigma_{\alpha} v^{\alpha}\right)_{; \mu}=\sigma_{\alpha ; \mu} v^{\alpha}+\sigma_{\alpha} v^{\alpha}{ }_{; \mu} \tag{10.22b}\\ & \left.\uparrow^{4}=\left(\sigma_{\alpha} v^{\alpha}\right)_{, \mu} \text { because } \sigma_{\alpha} v^{\alpha} \text { has no free indices to correct }\right] \end{align*}(10.22b)(σαvα);μ=σα;μvα+σαvα;μ4=(σαvα),μ because σαvα has no free indices to correct ]
(contract this with u μ u μ u^(mu)u^{\mu}uμ to get chain rule 10.7). Another example:
(10.22c) ( σ α ρ β v γ ) ; μ = σ α ; μ ρ β v γ + σ α ρ β ; μ v γ + σ α ρ β v γ ; μ (10.22c) σ α ρ β v γ ; μ = σ α ; μ ρ β v γ + σ α ρ β ; μ v γ + σ α ρ β v γ ; μ {:(10.22c)(sigma_(alpha)rho_(beta)v^(gamma))_(;mu)=sigma_(alpha;mu)rho_(beta)v^(gamma)+sigma_(alpha)rho_(beta;mu)v^(gamma)+sigma_(alpha)rho_(beta)v^(gamma)_(;mu):}\begin{equation*} \left(\sigma_{\alpha} \rho_{\beta} v^{\gamma}\right)_{; \mu}=\sigma_{\alpha ; \mu} \rho_{\beta} v^{\gamma}+\sigma_{\alpha} \rho_{\beta ; \mu} v^{\gamma}+\sigma_{\alpha} \rho_{\beta} v^{\gamma}{ }_{; \mu} \tag{10.22c} \end{equation*}(10.22c)(σαρβvγ);μ=σα;μρβvγ+σαρβ;μvγ+σαρβvγ;μ
(contract this with u μ u μ u^(mu)u^{\mu}uμ to get chain rule 10.11). Another example: see Exercise (10.12) below.

EXERCISES

Exercise 10.7. COMPUTATION OF CONNECTION COEFFICIENTS

Derive equation (10.14) for Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ from equation (10.13).
Exercise 10.8. CONNECTION FOR 1-FORM BASIS
Derive equations (10.15) and (10.16), which relate β ω ν β ω ν grad_(beta)omega^(nu)\boldsymbol{\nabla}_{\beta} \boldsymbol{\omega}^{\nu}βων to Γ ν α β Γ ν α β Gamma^(nu)_(alpha beta)\Gamma^{\nu}{ }_{\alpha \beta}Γναβ, from equation (10.14). Hint: use equation (10.7).

Exercise 10.9. SYMMETRY OF CONNECTION COEFFICIENTS

Show that the symmetry of spacetime's covariant derivative (equation 10.2a) is equivalent to the following symmetry condition on the connection coefficients:
(antisymmetric part of Γ μ α β ) 1 2 ( Γ μ α β Γ μ β α ) (10.23) Γ μ [ α β ] = 1 2 ω μ , [ e α , e β ] 1 2 c α β μ . [commutator of basis vectors] ]  (antisymmetric part of  Γ μ α β 1 2 Γ μ α β Γ μ β α (10.23) Γ μ [ α β ] = 1 2 ω μ , e α , e β 1 2 c α β μ .  [commutator of basis vectors]  ] {:[" (antisymmetric part of "{:Gamma^(mu)_(alpha beta))-=(1)/(2)(Gamma^(mu)_(alpha beta)-Gamma^(mu)_(beta alpha))],[(10.23)-=Gamma^(mu)_([alpha beta])=-(1)/(2)(:omega^(mu)","ubrace([e_(alpha),e_(beta)]ubrace)_(uarr):)-=-(1)/(2)c_(alpha beta)^(mu).],[" [commutator of basis vectors] "]]:}\begin{align*} & \text { (antisymmetric part of } \left.\Gamma^{\mu}{ }_{\alpha \beta}\right) \equiv \frac{1}{2}\left(\Gamma^{\mu}{ }_{\alpha \beta}-\Gamma^{\mu}{ }_{\beta \alpha}\right) \\ & \equiv \Gamma^{\mu}{ }_{[\alpha \beta]}=-\frac{1}{2}\langle\boldsymbol{\omega}^{\mu}, \underbrace{\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]}_{\uparrow}\rangle \equiv-\frac{1}{2} c_{\alpha \beta}{ }^{\mu} . \tag{10.23}\\ & \text { [commutator of basis vectors] }] \end{align*} (antisymmetric part of Γμαβ)12(ΓμαβΓμβα)(10.23)Γμ[αβ]=12ωμ,[eα,eβ]12cαβμ. [commutator of basis vectors] ]
As a special case, Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ is symmetric in α α alpha\alphaα and β β beta\betaβ when a coordinate basis ( e α = / x α ) e α = / x α (e_(alpha)=del//delx^(alpha))\left(\boldsymbol{e}_{\alpha}=\partial / \partial x^{\alpha}\right)(eα=/xα) is used. Show that in a coordinate basis this symmetry reduces the number of independent connection coefficients at each event from 4 × 4 × 4 = 64 4 × 4 × 4 = 64 4xx4xx4=644 \times 4 \times 4=644×4×4=64 to 4 × 10 = 40 4 × 10 = 40 4xx10=404 \times 10=404×10=40.

Exercise 10.10. COMPONENTS OF GRADIENT

Derive equation (10.18) for the components of the gradient, S α β γ ; 8 S α β γ ; 8 S^(alpha)_(beta gamma;8)S^{\alpha}{ }_{\beta \gamma ; 8}Sαβγ;8. Hint: Expand S S S\boldsymbol{S}S in terms of the given basis, and then evaluate the righthand side of
u S = u ( S α β γ e α ω β ω γ ) , u S = u S α β γ e α ω β ω γ , grad_(u)S=grad_(u)(S^(alpha)_(beta gamma)e_(alpha)oxomega^(beta)oxomega^(gamma)),\boldsymbol{\nabla}_{u} \boldsymbol{S}=\boldsymbol{\nabla}_{u}\left(S^{\alpha}{ }_{\beta \gamma} \boldsymbol{e}_{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma}\right),uS=u(Sαβγeαωβωγ),
for an arbitrary vector u u u\boldsymbol{u}u. Use the chain rules (10.2b) and (10.11). By comparing the result with
u S = S α β γ ; 8 u δ e α ω β ω γ , u S = S α β γ ; 8 u δ e α ω β ω γ , grad_(u)S=S^(alpha)_(beta gamma;8)u^(delta)e_(alpha)oxomega^(beta)oxomega^(gamma),\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{S}=S^{\alpha}{ }_{\beta \gamma ; 8} u^{\delta} \boldsymbol{e}_{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma},uS=Sαβγ;8uδeαωβωγ,
read off the components S α β γ ; δ S α β γ ; δ S^(alpha)_(beta gamma;delta)S^{\alpha}{ }_{\beta \gamma ; \delta}Sαβγ;δ.

Exercise 10.11. DIVERGENCE

Let T T T\boldsymbol{T}T be a ( 0 2 ) 0 0 2 0 _((_(0)^(2)))^(0){ }_{\left({ }_{0}^{2}\right)}^{0}(02)0 ) tensor field, and define the divergence on its second slot by the same process as in flat spacetime: T = T = grad*T=\boldsymbol{\nabla} \cdot \boldsymbol{T}=T= contraction of T T grad T\boldsymbol{\nabla} \boldsymbol{T}T; i.e.,
(10.24) ( T ) α = T α β ; β . (10.24) ( T ) α = T α β ; β . {:(10.24)(grad*T)^(alpha)=T^(alpha beta)_(;beta).:}\begin{equation*} (\boldsymbol{\nabla} \cdot \boldsymbol{T})^{\alpha}=T^{\alpha \beta}{ }_{; \beta} . \tag{10.24} \end{equation*}(10.24)(T)α=Tαβ;β.
Write the components T α β ; β T α β ; β T^(alpha beta)_(;beta)T^{\alpha \beta}{ }_{; \beta}Tαβ;β in terms of T α β β T α β β T^(alpha beta)_(beta)T^{\alpha \beta}{ }_{\beta}Tαββ plus correction terms for each of the two indices of T T T\boldsymbol{T}T.
[Answer:
T α β ; β = T α β , β + Γ α μ β T μ β + Γ β μ β T α μ . ] T α β ; β = T α β , β + Γ α μ β T μ β + Γ β μ β T α μ . ] T^(alpha beta)_(;beta)=T^(alpha beta)_(,beta)+Gamma^(alpha)_(mu beta)T^(mu beta)+Gamma^(beta)_(mu beta)T^(alpha mu)_(.])T^{\alpha \beta}{ }_{; \beta}=T^{\alpha \beta}{ }_{, \beta}+\Gamma^{\alpha}{ }_{\mu \beta} T^{\mu \beta}+\Gamma^{\beta}{ }_{\mu \beta} T^{\alpha \mu}{ }_{.]}Tαβ;β=Tαβ,β+ΓαμβTμβ+ΓβμβTαμ.]

Exercise 10.12. VERIFICATION OF Chain rule

Let S α β γ S α β γ S^(alpha beta)_(gamma)S^{\alpha \beta}{ }_{\gamma}Sαβγ be components of a ( ( 1 2 1 2 (_(1)^(2):}\left(_{1}^{2}\right.(12 ) tensor field, and M β γ M β γ M_(beta)^(gamma)M_{\beta}{ }^{\gamma}Mβγ be components of a ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11) tensor field. By contracting these tensor fields, one obtains a vector field S α β γ M β γ S α β γ M β γ S^(alpha beta)_(gamma)M_(beta)^(gamma)S^{\alpha \beta}{ }_{\gamma} M_{\beta}{ }^{\gamma}SαβγMβγ. The chain rule for the divergence of this vector field reads
( S α β γ M β γ ) ; α = S α β γ ; α M β γ + S α β γ M β γ ; α . S α β γ M β γ ; α = S α β γ ; α M β γ + S α β γ M β γ ; α . (S^(alpha beta)_(gamma)M_(beta)^(gamma))_(;alpha)=S^(alpha beta)_(gamma;alpha)M_(beta)^(gamma)+S^(alpha beta)_(gamma)M_(beta)^(gamma)_(;alpha).\left(S^{\alpha \beta}{ }_{\gamma} M_{\beta}{ }^{\gamma}\right)_{; \alpha}=S^{\alpha \beta}{ }_{\gamma ; \alpha} M_{\beta}{ }^{\gamma}+S^{\alpha \beta}{ }_{\gamma} M_{\beta}{ }^{\gamma}{ }_{; \alpha} .(SαβγMβγ);α=Sαβγ;αMβγ+SαβγMβγ;α.
Verify the validity of this chain rule by expressing both sides of the equation in terms of directional derivatives ( e e e e e_(e)\boldsymbol{e}_{\boldsymbol{e}}ee ) plus connection-coefficient corrections. Hint: the left side becomes
The right side has many more correction terms (three on S α β γ ; α S α β γ ; α S^(alpha beta)_(gamma;alpha)S^{\alpha \beta}{ }_{\gamma ; \alpha}Sαβγ;α; two on M β γ ; α M β γ ; α M_(beta)^(gamma);alphaM_{\beta}{ }^{\gamma} ; \alphaMβγ;α ), but they must cancel against each other, leaving only one.

Exercise 10.13. TRANSFORMATION LAW FOR CONNECTION COEFFICIENTS

Let { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} and { e μ } e μ {e_(mu^('))}\left\{\boldsymbol{e}_{\mu^{\prime}}\right\}{eμ} be two different fields of basis vectors related by the transformation law
(10.25) e μ ( P ) = L α μ ( P ) e α ( P ) . (10.25) e μ ( P ) = L α μ ( P ) e α ( P ) . {:(10.25)e_(mu^('))(P)=L^(alpha)_(mu^('))(P)e_(alpha)(P).:}\begin{equation*} \boldsymbol{e}_{\mu^{\prime}}(\mathscr{P})=L^{\alpha}{ }_{\mu^{\prime}}(\mathscr{P}) \boldsymbol{e}_{\alpha}(\mathscr{P}) . \tag{10.25} \end{equation*}(10.25)eμ(P)=Lαμ(P)eα(P).
Show that the corresponding connection coefficients are related by
(10.26) Γ α β γ = L α L μ β L ν γ Γ ρ μ ν + L α L μ β , γ (10.26) Γ α β γ = L α L μ β L ν γ Γ ρ μ ν + L α L μ β , γ {:(10.26)Gamma^(alpha^('))_(beta^(')gamma^('))=ubrace(L^(alpha^('))L^(mu)_(beta)^(')L^(nu)_(gamma^('))Gamma^(rho)_(mu nu)ubrace)+L^(alpha^('))L^(mu)_(beta^('),gamma^(')):}\begin{equation*} \Gamma^{\alpha^{\prime}}{ }_{\beta^{\prime} \gamma^{\prime}}=\underbrace{L^{\alpha^{\prime}} L^{\mu}{ }_{\beta}{ }^{\prime} L^{\nu}{ }_{\gamma^{\prime}} \Gamma^{\rho}{ }_{\mu \nu}}+L^{\alpha^{\prime}} L^{\mu}{ }_{\beta^{\prime}, \gamma^{\prime}} \tag{10.26} \end{equation*}(10.26)Γαβγ=LαLμβLνγΓρμν+LαLμβ,γ
standard transformation law for components of a tensor

Exercise 10.14. POLAR COORDINATES IN FLAT 2-DIMENSIONAL SPACE

On a sheet of paper draw an ( r , ϕ ) ( r , ϕ ) (r,phi)(r, \phi)(r,ϕ) polar coordinate system. At neighboring points, draw the basis vectors e r ^ = / r e r ^ = / r e_( hat(r))=del//del r\boldsymbol{e}_{\hat{r}}=\partial / \partial rer^=/r and e ϕ ^ r 1 / ϕ ˙ e ϕ ^ r 1 / ϕ ˙ e_( hat(phi))-=r^(-1)del//delphi^(˙)\boldsymbol{e}_{\hat{\phi}} \equiv r^{-1} \partial / \partial \dot{\phi}eϕ^r1/ϕ˙. (a) Use this picture, and Euclid's version of parallel transport, to justify the relations
r ^ e r ^ = 0 , r ^ e δ ^ = 0 , ϕ ^ e r ^ = r 1 e ϕ ^ , ζ ^ e ϕ ^ = r 1 e r ^ r ^ e r ^ = 0 , r ^ e δ ^ = 0 , ϕ ^ e r ^ = r 1 e ϕ ^ , ζ ^ e ϕ ^ = r 1 e r ^ grad_( hat(r))e_( hat(r))=0,quadgrad_( hat(r))e_( hat(delta))=0,quadgrad_( hat(phi))e_( hat(r))=r^(-1)e_( hat(phi)),quadgrad_( hat(zeta))e_( hat(phi))=-r^(-1)e_( hat(r))\boldsymbol{\nabla}_{\hat{r}} \boldsymbol{e}_{\hat{r}}=0, \quad \boldsymbol{\nabla}_{\hat{r}} \boldsymbol{e}_{\hat{\delta}}=0, \quad \boldsymbol{\nabla}_{\hat{\phi}} \boldsymbol{e}_{\hat{r}}=r^{-1} \boldsymbol{e}_{\hat{\phi}}, \quad \boldsymbol{\nabla}_{\hat{\zeta}} \boldsymbol{e}_{\hat{\phi}}=-r^{-1} \boldsymbol{e}_{\hat{r}}r^er^=0,r^eδ^=0,ϕ^er^=r1eϕ^,ζ^eϕ^=r1er^
(b) From these relations write down the connection coefficients. (c) Let A = A r ^ e r ^ + A ϕ ^ e ϕ ^ A = A r ^ e r ^ + A ϕ ^ e ϕ ^ A=A^( hat(r))e_( hat(r))+A^( hat(phi))e_( hat(phi))\boldsymbol{A}=A^{\hat{r}} \boldsymbol{e}_{\hat{r}}+A^{\hat{\phi}} \boldsymbol{e}_{\hat{\phi}}A=Ar^er^+Aϕ^eϕ^ be a vector field. Show that its divergence, A = A α ^ ; α ^ = A α ^ , α ^ + Γ α ^ μ ^ α ^ A μ ^ A = A α ^ ; α ^ = A α ^ , α ^ + Γ α ^ μ ^ α ^ A μ ^ grad*A=A^( hat(alpha))_(; hat(alpha))=A^( hat(alpha))_(, hat(alpha))+Gamma^( hat(alpha))_( hat(mu) hat(alpha))A^( hat(mu))\boldsymbol{\nabla} \cdot \boldsymbol{A}=A^{\hat{\alpha}}{ }_{; \hat{\alpha}}=A^{\hat{\alpha}}{ }_{, \hat{\alpha}}+\Gamma^{\hat{\alpha}}{ }_{\hat{\mu} \hat{\alpha}} A^{\hat{\mu}}A=Aα^;α^=Aα^,α^+Γα^μ^α^Aμ^, can be calculated using the formula
A = 1 r A ϕ ^ ϕ + 1 r ( r A r ^ ) r A = 1 r A ϕ ^ ϕ + 1 r r A r ^ r grad*A=(1)/(r)(delA^( hat(phi)))/(del phi)+(1)/(r)(del(rA^( hat(r))))/(del r)\boldsymbol{\nabla} \cdot \boldsymbol{A}=\frac{1}{r} \frac{\partial A^{\hat{\phi}}}{\partial \phi}+\frac{1}{r} \frac{\partial\left(r A^{\hat{r}}\right)}{\partial r}A=1rAϕ^ϕ+1r(rAr^)r
(which should be familiar to most readers).
Geodesic equation: abstract version

§10.5. GEODESIC EQUATION

Geodesics-the parametrized paths of freely falling particles-were the starting point of this chapter. From them parallel transport was constructed (Schild's ladder; Box 10.2 ); and parallel transport in turn produced the covariant derivative and its connection coefficients. Given the covariant derivative, one recovered the geodesics: they were the curves whose tangent vectors, u = d P / d λ u = d P / d λ u=dP//d lambda\boldsymbol{u}=d \mathscr{P} / d \lambdau=dP/dλ, satisfy u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 ( u u u\boldsymbol{u}u is parallel transported along itself).
Let a coordinate system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)} be given. Let it induce basis vectors e α = / x α e α = / x α e_(alpha)=del//delx^(alpha)\boldsymbol{e}_{\alpha}=\partial / \partial x^{\alpha}eα=/xα into the tangent space at each event. Let the connection coefficients Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ for this "coordinate basis" be given. Then the component version of the "geodesic equation" u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 becomes a differential equation for the geodesic x α ( λ ) x α ( λ ) x^(alpha)(lambda)x^{\alpha}(\lambda)xα(λ) :
(1) u = d d λ = d x α d λ x α components of u are u α = d x α d λ (1) u = d d λ = d x α d λ x α  components of  u  are  u α = d x α d λ {:(1)u=(d)/(d lambda)=(dx^(alpha))/(d lambda)(del)/(delx^(alpha))quad Longrightarrowquad" components of "u" are "u^(alpha)=(dx^(alpha))/(d lambda):}\begin{equation*} \boldsymbol{u}=\frac{d}{d \lambda}=\frac{d x^{\alpha}}{d \lambda} \frac{\partial}{\partial x^{\alpha}} \quad \Longrightarrow \quad \text { components of } \boldsymbol{u} \text { are } u^{\alpha}=\frac{d x^{\alpha}}{d \lambda} \tag{1} \end{equation*}(1)u=ddλ=dxαdλxα components of u are uα=dxαdλ
(2) then components of u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 are
0 = u α ; β u β = ( u α , β + Γ α γ β u γ ) u β = x β ( d x α d λ ) d x β d λ + Γ α γ β d x γ d λ d x β d λ , 0 = u α ; β u β = u α , β + Γ α γ β u γ u β = x β d x α d λ d x β d λ + Γ α γ β d x γ d λ d x β d λ , {:[0=u^(alpha)_(;beta)u^(beta)=(u^(alpha)_(,beta)+Gamma^(alpha)_(gamma beta)u^(gamma))u^(beta)],[=(del)/(delx^(beta))((dx^(alpha))/(d lambda))(dx^(beta))/(d lambda)+Gamma^(alpha)_(gamma beta)(dx^(gamma))/(d lambda)(dx^(beta))/(d lambda)","]:}\begin{aligned} 0 & =u^{\alpha}{ }_{; \beta} u^{\beta}=\left(u^{\alpha}{ }_{, \beta}+\Gamma^{\alpha}{ }_{\gamma \beta} u^{\gamma}\right) u^{\beta} \\ & =\frac{\partial}{\partial x^{\beta}}\left(\frac{d x^{\alpha}}{d \lambda}\right) \frac{d x^{\beta}}{d \lambda}+\Gamma^{\alpha}{ }_{\gamma \beta} \frac{d x^{\gamma}}{d \lambda} \frac{d x^{\beta}}{d \lambda}, \end{aligned}0=uα;βuβ=(uα,β+Γαγβuγ)uβ=xβ(dxαdλ)dxβdλ+Γαγβdxγdλdxβdλ,
which reduces to the differential equation
(10.27) d 2 x α d λ 2 + Γ α γ β d x γ d λ d x β d λ = 0 . (10.27) d 2 x α d λ 2 + Γ α γ β d x γ d λ d x β d λ = 0 . {:(10.27)(d^(2)x^(alpha))/(dlambda^(2))+Gamma^(alpha)_(gamma beta)(dx^(gamma))/(d lambda)(dx^(beta))/(d lambda)=0.:}\begin{equation*} \frac{d^{2} x^{\alpha}}{d \lambda^{2}}+\Gamma^{\alpha}{ }_{\gamma \beta} \frac{d x^{\gamma}}{d \lambda} \frac{d x^{\beta}}{d \lambda}=0 . \tag{10.27} \end{equation*}(10.27)d2xαdλ2+Γαγβdxγdλdxβdλ=0.
This component version of the geodesic equation gives an analytic method ("translation" of Schild's ladder) for constructing the parallel transport law from a knowledge of the geodesics. Pick an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 and set up a coordinate system in its neighborhood. Watch many clock-carrying particles pass through (or arbitrarily close to) P 0 P 0 P_(0)\mathscr{P}_{0}P0. For each particle read off the values of d 2 x α / d λ 2 d 2 x α / d λ 2 d^(2)x^(alpha)//dlambda^(2)d^{2} x^{\alpha} / d \lambda^{2}d2xα/dλ2 and d x α / d λ d x α / d λ dx^(alpha)//d lambdad x^{\alpha} / d \lambdadxα/dλ at P 0 P 0 P_(0)\mathscr{P}_{0}P0. Insert all the data for many particles into equation (10.27), and solve for the connection coefficients. Do not be disturbed that only the symmetric part of Γ α γ β Γ α γ β Gamma^(alpha)_(gamma beta)\Gamma^{\alpha}{ }_{\gamma \beta}Γαγβ is obtained thereby; the antisymmetric part, Γ [ γ β ] α Γ [ γ β ] α Gamma_([gamma beta])^(alpha)\Gamma_{[\gamma \beta]}^{\alpha}Γ[γβ]α, vanishes identically in any coordinate frame! (See exercise 10.9.) Knowing Γ α γ β Γ α γ β Gamma^(alpha)_(gamma beta)\Gamma^{\alpha}{ }_{\gamma \beta}Γαγβ, use them to parallel transport any desired vector along any desired curve through P 0 P 0 P_(0)\mathscr{P}_{0}P0 :
(10.28) u v = 0 d v α d λ + Γ α γ β v γ d x β d λ = 0 (10.28) u v = 0 d v α d λ + Γ α γ β v γ d x β d λ = 0 {:(10.28)grad_(u)v=0quad Longleftrightarrowquad(dv^(alpha))/(d lambda)+Gamma^(alpha)_(gamma beta)v^(gamma)(dx^(beta))/(d lambda)=0:}\begin{equation*} \boldsymbol{\nabla}_{u} \boldsymbol{v}=0 \quad \Longleftrightarrow \quad \frac{d v^{\alpha}}{d \lambda}+\Gamma^{\alpha}{ }_{\gamma \beta} v^{\gamma} \frac{d x^{\beta}}{d \lambda}=0 \tag{10.28} \end{equation*}(10.28)uv=0dvαdλ+Γαγβvγdxβdλ=0

Exercise 10.15. COMPONENTS OF PARALLEL-TRANSPORT LAW

EXERCISES

Show that equation (10.28) is the component version of the law for parallel transporting a vector v v v\boldsymbol{v}v along the curve P ( λ ) P ( λ ) P(lambda)\mathscr{\mathscr { P }}(\lambda)P(λ) with tangent vector u = d P / d λ u = d P / d λ u=dP//d lambda\boldsymbol{u}=d \mathscr{P} / d \lambdau=dP/dλ.

Exercise 10.16. GEODESICS IN POLAR COORDINATES

In rectangular coordinates on a flat sheet of paper, Euclid's straight lines (geodesics) satisfy d 2 x / d λ 2 = d 2 y / d λ 2 = 0 d 2 x / d λ 2 = d 2 y / d λ 2 = 0 d^(2)x//dlambda^(2)=d^(2)y//dlambda^(2)=0d^{2} x / d \lambda^{2}=d^{2} y / d \lambda^{2}=0d2x/dλ2=d2y/dλ2=0. Transform this geodesic equation into polar coordinates ( x = ( x = (x=(x=(x= r cos ϕ , y = r sin ϕ r cos ϕ , y = r sin ϕ r cos phi,y=r sin phir \cos \phi, y=r \sin \phircosϕ,y=rsinϕ ); and read off the resulting connection coefficients by comparison with equation (10.27). These are the connection coefficients for the coordinate basis ( / r ( / r (del//del r(\partial / \partial r(/r, / ϕ ) / ϕ ) del//del phi)\partial / \partial \phi)/ϕ). From them calculate the connection coefficients for the basis
e r ^ = r , e ϕ ^ = 1 r ϕ ˙ . e r ^ = r , e ϕ ^ = 1 r ϕ ˙ . e_( hat(r))=(del)/(del r),quade_( hat(phi))=(1)/(r)(del)/(del(phi^(˙))).\boldsymbol{e}_{\hat{r}}=\frac{\partial}{\partial r}, \quad \boldsymbol{e}_{\hat{\phi}}=\frac{1}{r} \frac{\partial}{\partial \dot{\phi}} .er^=r,eϕ^=1rϕ˙.
The answer should agree with the answer to part (b) of Exercise 10.14. Hint: Use such relations as
e β ^ e r ^ = ( 1 / r ( / ϕ ) ( / r ) = 1 r ( / ϕ ) ( / r ) . e β ^ e r ^ = ( 1 / r ( / ϕ ) ( / r ) = 1 r ( / ϕ ) ( / r ) . grad_(e_( hat(beta)))e_( hat(r))=grad_((1//r del(del//del phi))(del//del r)=(1)/(r)grad_((del//del phi))(del//del r).\boldsymbol{\nabla}_{\boldsymbol{e}_{\hat{\beta}}} \boldsymbol{e}_{\hat{r}}=\boldsymbol{\nabla}_{(1 / r \partial(\partial / \partial \phi)}(\partial / \partial r)=\frac{1}{r} \boldsymbol{\nabla}_{(\partial / \partial \phi)}(\partial / \partial r) .eβ^er^=(1/r(/ϕ)(/r)=1r(/ϕ)(/r).
How to construct parallel transport law from knowledge of geodesics

Exercise 10.17. ROTATION GROUP: GEODESICS AND CONNECTION COEFFICIENTS

[Continuation of exercises 9.13 and 9.14.] In discussing the rotation group, one must make a clear distinction between the Euclidean space (coordinates x , y , z x , y , z x,y,zx, y, zx,y,z; basis vectors / x , / y / x , / y del//del x,del//del y\partial / \partial x, \partial / \partial y/x,/y, / z ) / z ) del//del z)\partial / \partial z)/z) in which the rotation matrices act, and the group manifold S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) (coordinates ψ , θ ψ , θ psi,theta\psi, \thetaψ,θ, ϕ ϕ phi\phiϕ; coordinate basis / ψ , / θ , / ϕ / ψ , / θ , / ϕ del//del psi,del//del theta,del//del phi\partial / \partial \psi, \partial / \partial \theta, \partial / \partial \phi/ψ,/θ,/ϕ; basis of "generators" e 1 , e 2 , e 3 e 1 , e 2 , e 3 e_(1),e_(2),e_(3)\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}e1,e2,e3 ), whose points P P P\mathscr{P}P are rotation matrices.
(a) Pick a vector
n = n x / x + n y / y + n z / z n = n x / x + n y / y + n z / z n=n^(x)del//del x+n^(y)del//del y+n^(z)del//del z\boldsymbol{n}=n^{x} \partial / \partial x+n^{y} \partial / \partial y+n^{z} \partial / \partial zn=nx/x+ny/y+nz/z
in Euclidean space. Show that
(10.29) n ( t ) exp [ ( n x K 1 + n y K 2 + n z K 3 ) t ] (10.29) n ( t ) exp n x K 1 + n y K 2 + n z K 3 t {:(10.29)ℜ_(n)(t)-=exp[(n^(x)K_(1)+n^(y)K_(2)+n^(z)K_(3))t]:}\begin{equation*} \mathscr{\Re}_{n}(t) \equiv \exp \left[\left(n^{x} \mathscr{K}_{1}+n^{y} \mathscr{K}_{2}+n^{z} \mathscr{K}_{3}\right) t\right] \tag{10.29} \end{equation*}(10.29)n(t)exp[(nxK1+nyK2+nzK3)t]
is a rotation matrix that rotates the axes of Euclidean space by an angle
t | n | = t [ ( n x ) 2 + ( n y ) 2 + ( n z ) 2 ] 1 / 2 t | n | = t n x 2 + n y 2 + n z 2 1 / 2 t|n|=t[(n^(x))^(2)+(n^(y))^(2)+(n^(z))^(2)]^(1//2)t|\boldsymbol{n}|=t\left[\left(n^{x}\right)^{2}+\left(n^{y}\right)^{2}+\left(n^{z}\right)^{2}\right]^{1 / 2}t|n|=t[(nx)2+(ny)2+(nz)2]1/2
about the direction n n n\boldsymbol{n}n. ( H j H j (H_(j):}\left(\mathscr{H}_{j}\right.(Hj are matrices defined in exercise 9.13.)
(b) In the group manifold S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3), pick a point (rotation matrix) P P P\mathscr{P}P, and pick a tangent vector u = u α e α u = u α e α u=u^(alpha)e_(alpha)\boldsymbol{u}=u^{\alpha} \boldsymbol{e}_{\alpha}u=uαeα at P P P\mathscr{P}P. Let u u u\boldsymbol{u}u be a vector in Euclidean space with the same components as u u u\boldsymbol{u}u has in S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) :
(10.30) u = u 1 e 1 + u 2 e 2 + u 3 e 3 ; u = u 1 / x + u 2 / y + u 3 / z . (10.30) u = u 1 e 1 + u 2 e 2 + u 3 e 3 ; u = u 1 / x + u 2 / y + u 3 / z . {:(10.30)u=u^(1)e_(1)+u^(2)e_(2)+u^(3)e_(3);quad u=u^(1)del//del x+u^(2)del//del y+u^(3)del//del z.:}\begin{equation*} \boldsymbol{u}=u^{1} \boldsymbol{e}_{1}+u^{2} \boldsymbol{e}_{2}+u^{3} \boldsymbol{e}_{3} ; \quad \boldsymbol{u}=u^{1} \partial / \partial x+u^{2} \partial / \partial y+u^{3} \partial / \partial z . \tag{10.30} \end{equation*}(10.30)u=u1e1+u2e2+u3e3;u=u1/x+u2/y+u3/z.
Show that u u u\boldsymbol{u}u is the tangent vector (at t = 0 t = 0 t=0t=0t=0 ) to the curve
(10.31) C ( t ) = R u ( t ) P . (10.31) C ( t ) = R u ( t ) P . {:(10.31)C(t)=R_(u)(t)P.:}\begin{equation*} \mathcal{C}(t)=\mathscr{R}_{u}(t) \mathscr{P} . \tag{10.31} \end{equation*}(10.31)C(t)=Ru(t)P.
The curve E ( t ) E ( t ) E(t)\mathcal{E}(t)E(t) through the arbitrary point P P P\mathscr{P}P with arbitrary tangent vector u = ( d C / d t ) t = 0 u = ( d C / d t ) t = 0 u=(dC//dt)_(t=0)\boldsymbol{u}=(d \mathcal{C} / d t)_{t=0}u=(dC/dt)t=0 is a very special curve: every point on it differs from P P P\mathscr{P}P by a rotation R u ( t ) R u ( t ) R_(u)(t)\mathscr{R}_{u}(t)Ru(t) about one and the same direction u u u\boldsymbol{u}u. No other curve in S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) with "starting conditions" { P , u } { P , u } {P,u}\{\mathscr{P}, \boldsymbol{u}\}{P,u} has such beautiful simplicity. Hence it is natural to decree that each such E ( t ) E ( t ) E(t)\mathcal{E}(t)E(t) is a geodesic of the group manifold S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3). This decree adds new geometric structure to S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3); it converts S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) from a differentiable manifold into something more special: an affine manifold.
One has no guarantee that an arbitrarily chosen family of curves in an arbitrary manifold can be decreed to be geodesics. Most families of curves simply do not possess the right geometric properties to function as geodesics. Most will lead to covariant derivatives that violate one or more of the fundamental conditions (10.2). To learn whether a given choice of geodesics is possible, one can try to derive connection coefficients Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ (for some given basis) corresponding to the chosen geodesics. If the derivation is successful, the choice of geodesics was a possible one. If the derivation produces inconsistencies, the chosen family of curves have the wrong geometric properties to function as geodesics.
(c) For the basis of generators { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} derive connection coefficients corresponding to the chosen geodesics, E ( t ) = R u ( t ) P E ( t ) = R u ( t ) P E(t)=R_(u)(t)P\mathcal{E}(t)=\mathscr{R}_{u}(t) \mathscr{P}E(t)=Ru(t)P, of S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3). Hint: show that the components u α = ω α , u u α = ω α , u u^(alpha)=(:omega^(alpha),u:)u^{\alpha}=\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{u}\right\rangleuα=ωα,u of the tangent u = d C / d t u = d C / d t u=dC//dt\boldsymbol{u}=d \mathcal{C} / d tu=dC/dt to a given geodesic are independent of position C ( t ) C ( t ) C(t)\mathcal{C}(t)C(t) along the geodesic. Then use the geodesic equation u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0, expanded in the basis { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα}, to calculate the symmetric part of the connection Γ α ( β γ ) Γ α ( β γ ) Gamma^(alpha)_((beta gamma))\Gamma^{\alpha}{ }_{(\beta \gamma)}Γα(βγ). Finally use equation (10.23) to calculate Γ α [ β γ ] Γ α [ β γ ] Gamma^(alpha)_([beta gamma])\Gamma^{\alpha}{ }_{[\beta \gamma]}Γα[βγ]. [Answer:
(10.32) Γ β γ α = 1 2 ϵ α β γ , (10.32) Γ β γ α = 1 2 ϵ α β γ , {:(10.32)Gamma_(beta gamma)^(alpha)=(1)/(2)epsilon_(alpha beta gamma)",":}\begin{equation*} \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2} \epsilon_{\alpha \beta \gamma}, \tag{10.32} \end{equation*}(10.32)Γβγα=12ϵαβγ,
where ϵ α β γ ϵ α β γ epsilon_(alpha beta gamma)\epsilon_{\alpha \beta \gamma}ϵαβγ is the completely antisymmetric symbol with ϵ 123 = + 1 ϵ 123 = + 1 epsilon_(123)=+1\epsilon_{123}=+1ϵ123=+1. This answer is independent of location P P P\mathscr{P}P in S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) !]

GEODESIC DEVIATION AND SPACETIME CURVATURE

§11.1. CURVATURE, AT LAST!

Spacetime curvature manifests itself as gravitation, by means of the deviation of one geodesic from a nearby geodesic (relative acceleration of test particles).
Let the geodesics of spacetime be known. Then the covariant derivative grad\boldsymbol{\nabla} and its connection coefficients Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ are also known. How, from this information, does one define, calculate, and understand geodesic deviation and spacetime curvature? The answer unfolds in this chapter, and is summarized in Box 11.1. To disclose the answer one must (1) define the "relative acceleration vector" u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} \boldsymbol{n}uun, which measures the deviation of one geodesic from another (§11.2); (2) derive an expression in terms of grad\boldsymbol{\nabla} or Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ for the "Riemann curvature tensor," which produces the geodesic deviation (§11.3); (3) see Riemann curvature at work, producing changes in vectors that are parallel transported around closed circuits ($11.4); (4) see Riemann curvature test whether spacetime is flat ( $ 11.5 $ 11.5 $11.5\$ 11.5$11.5 ); and (5) construct a special coordinate system, "Riemann normal coordinates," which is tied in a special way to the Riemann curvature tensor (§11.6).

§11.2. THE RELATIVE ACCELERATION OF NEIGHBORING GEODESICS

Focus attention on a family of geodesics (Figure 11.1). Let one geodesic be distinguished from another by the value of a "selector parameter" n n nnn. The family includes not only geodesics n = 0 , 1 , 2 , n = 0 , 1 , 2 , n=0,1,2,dotsn=0,1,2, \ldotsn=0,1,2, but also geodesics for all intervening values of
This chapter is entirely Track 2. Chapters 9 and 10 are necessary preparation for it.
It will be needed as preparation for
(1) Chapters 12 and 13 (Newtonian gravity; Riemannian geometry),
(2) the second half of Chapter 14 (calculation of curvature), and
(3) the details, but not the message, of Chapter 15 (Bianchi identities).
Overview of chapter

Box 11.1 GEODESIC DEVIATION AND RIEMANN CURVATURE IN BRIEF

"Geodesic separation" n n n\boldsymbol{n}n is displacement (tangent vector) from point on fiducial geodesic to point on nearby geodesic characterized by same value of affine parameter λ λ lambda\lambdaλ.
Geodesic separation changes with respect to λ λ lambda\lambdaλ (i.e., changes along the tangent vector u = d / d λ u = d / d λ u=d//d lambda\boldsymbol{u}=d / d \lambdau=d/dλ ) at a rate given by the equation of geodesic deviation
(1) u u n + Riemann ( , u , n , u ) = 0 (1) u u n +  Riemann  ( , u , n , u ) = 0 {:(1)grad_(u)grad_(u)n+" Riemann "(dots","u","n","u)=0:}\begin{equation*} \nabla_{u} \nabla_{u} n+\text { Riemann }(\ldots, u, n, u)=0 \tag{1} \end{equation*}(1)uun+ Riemann (,u,n,u)=0
(second-order equation; see § § 1.6 § § 1.6 §§1.6\S \S 1.6§§1.6 and 1.7; Figures 1.10, 1.11, 1.12).
In terms of components of the Riemann tensor the driving force ("tidal graviational force") is
Riemann ( , u , n , u ) = e α R α β γ δ u β n γ u δ ( , u , n , u ) = e α R α β γ δ u β n γ u δ (dots,u,n,u)=e_(alpha)R^(alpha)_(beta gamma delta)u^(beta)n^(gamma)u^(delta)(\ldots, \boldsymbol{u}, \boldsymbol{n}, \boldsymbol{u})=\boldsymbol{e}_{\alpha} R^{\alpha}{ }_{\beta \gamma \delta} u^{\beta} n^{\gamma} u^{\delta}(,u,n,u)=eαRαβγδuβnγuδ.
The components of the Riemann curvature tensor in a coordinate frame are given in terms of the connection coefficients by the formula
R β γ δ α = Γ β δ α x γ Γ β γ α x δ (3) + Γ μ γ β γ Γ β δ μ Γ μ δ α Γ β γ μ R β γ δ α = Γ β δ α x γ Γ β γ α x δ (3) + Γ μ γ β γ Γ β δ μ Γ μ δ α Γ β γ μ {:[R_(beta gamma delta)^(alpha)=(delGamma_(beta delta)^(alpha))/(delx^(gamma))-(delGamma_(beta gamma)^(alpha))/(delx^(delta))],[(3)+Gamma^(mu gamma)_(beta gamma)Gamma_(beta delta)^(mu)-Gamma_(mu delta)^(alpha)Gamma_(beta gamma)^(mu)]:}\begin{align*} R_{\beta \gamma \delta}^{\alpha}= & \frac{\partial \Gamma_{\beta \delta}^{\alpha}}{\partial x^{\gamma}}-\frac{\partial \Gamma_{\beta \gamma}^{\alpha}}{\partial x^{\delta}} \\ & +\Gamma^{\mu \gamma}{ }_{\beta \gamma} \Gamma_{\beta \delta}^{\mu}-\Gamma_{\mu \delta}^{\alpha} \Gamma_{\beta \gamma}^{\mu} \tag{3} \end{align*}Rβγδα=ΓβδαxγΓβγαxδ(3)+ΓμγβγΓβδμΓμδαΓβγμ
This curvature tensor not only quantifies the concept of "tidal gravitational force," but also enters into Einstein's law, by which "matter tells spacetime how to curve." That law, to be studied
in later chapters, takes the following operationalcomputational form in a given coordinate system:
(a) Write down trial formula for dynamic evolution of metric coefficients g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν with time.
(b) Calculate the connection coefficients from
(4) Γ α μ ν = g α β Γ β μ ν ; (5) Γ β μ ν = 1 2 ( g β ν x μ + g β μ x ν g μ ν x β ) (4) Γ α μ ν = g α β Γ β μ ν ; (5) Γ β μ ν = 1 2 g β ν x μ + g β μ x ν g μ ν x β {:[(4)Gamma^(alpha)_(mu nu)=g^(alpha beta)Gamma_(beta mu nu);],[(5)Gamma_(beta mu nu)=(1)/(2)((delg_(beta nu))/(delx^(mu))+(delg_(beta mu))/(delx^(nu))-(delg_(mu nu))/(delx^(beta)))]:}\begin{gather*} \Gamma^{\alpha}{ }_{\mu \nu}=g^{\alpha \beta} \Gamma_{\beta \mu \nu} ; \tag{4}\\ \Gamma_{\beta \mu \nu}=\frac{1}{2}\left(\frac{\partial g_{\beta \nu}}{\partial x^{\mu}}+\frac{\partial g_{\beta \mu}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\beta}}\right) \tag{5} \end{gather*}(4)Γαμν=gαβΓβμν;(5)Γβμν=12(gβνxμ+gβμxνgμνxβ)
(derived in Chapter 13).
(c) Calculate Riemann curvature tensor from equation (3).
(d) Calculate Einstein curvature tensor from
(6) G μ ν = R μ α ν α 1 2 g μ ν g σ τ R σ α τ α (6) G μ ν = R μ α ν α 1 2 g μ ν g σ τ R σ α τ α {:(6)G_(mu nu)=R_(mu alpha nu)^(alpha)-(1)/(2)g_(mu nu)g^(sigma tau)R_(sigma alpha tau)^(alpha):}\begin{equation*} G_{\mu \nu}=R_{\mu \alpha \nu}^{\alpha}-\frac{1}{2} g_{\mu \nu} g^{\sigma \tau} R_{\sigma \alpha \tau}^{\alpha} \tag{6} \end{equation*}(6)Gμν=Rμανα12gμνgστRσατα
(geometric significance in Chapter 15).
(e) Insert into Einstein's equations (Chapter 17):
G μ ν = 0 (empty space) G μ ν = 8 π T μ ν (when mass-energy is present). G μ ν = 0  (empty space)  G μ ν = 8 π T μ ν  (when mass-energy is   present).  {:[G_(mu nu)=0," (empty space) "],[G_(mu nu)=8piT_(mu nu)," (when mass-energy is "],[" present). "]:}\begin{array}{ll} G_{\mu \nu}=0 & \text { (empty space) } \\ G_{\mu \nu}=8 \pi T_{\mu \nu} & \text { (when mass-energy is } \\ \text { present). } \end{array}Gμν=0 (empty space) Gμν=8πTμν (when mass-energy is  present). 
(f) Test whether the trial formula for the dynamic evolution of the geometry was correct, and, if not, change it so it is.
Affine parameter
n n nnn. The typical point P P P\mathscr{P}P on the typical geodesic will be a continuous, doubly differentiable function of the selector parameter n n nnn and the affine parameter λ λ lambda\lambdaλ; thus
(11.1) P = P ( λ , n ) . (11.1) P = P ( λ , n ) . {:(11.1)P=P(lambda","n).:}\begin{equation*} \mathscr{P}=\mathscr{P}(\lambda, n) . \tag{11.1} \end{equation*}(11.1)P=P(λ,n).
The tangent vector
u = P λ (Cartan notation) u = P λ  (Cartan notation)  u=(delP)/(del lambda)quad" (Cartan notation) "\boldsymbol{u}=\frac{\partial \mathscr{P}}{\partial \lambda} \quad \text { (Cartan notation) }u=Pλ (Cartan notation) 
or
(11.2) u = λ (notation of this book) (11.2) u = λ  (notation of this book)  {:(11.2)u=(del)/(del lambda)quad" (notation of this book) ":}\begin{equation*} \boldsymbol{u}=\frac{\partial}{\partial \lambda} \quad \text { (notation of this book) } \tag{11.2} \end{equation*}(11.2)u=λ (notation of this book) 
is constant along any given geodesic in this sense: the vector u u u\boldsymbol{u}u at any point, trans-
Figure 11.1
One-parameter family of geodesics. The "selector parameter" n n nnn tells which geodesic. The affine parameter λ λ lambda\lambdaλ tells where on a given geodesic. The two tangent vectors indicated in the diagram are u = u = u=\boldsymbol{u}=u= / λ ( / λ ( del//del lambda(\partial / \partial \lambda(/λ( Cartan: P / λ ) P / λ ) delP//del lambda)\partial \mathscr{P} / \partial \lambda)P/λ) and n = / n ( n = / n ( n=del//del n(\boldsymbol{n}=\partial / \partial n(n=/n( Cartan: P / n ) P / n ) delP//del n)\partial \mathscr{P} / \partial n)P/n).
ported parallel to itself along the geodesic, arrives at a second point coincident in direction and length with the u u u\boldsymbol{u}u already existing at that point.
The "separation vector"
n = P n (Cartan notation) n = P n  (Cartan notation)  n=(delP)/(del n)quad" (Cartan notation) "\boldsymbol{n}=\frac{\partial \mathscr{P}}{\partial n} \quad \text { (Cartan notation) }n=Pn (Cartan notation) 
or
(11.3) n = n (notation of this book) (11.3) n = n  (notation of this book)  {:(11.3)n=(del)/(del n)quad" (notation of this book) ":}\begin{equation*} \boldsymbol{n}=\frac{\partial}{\partial \boldsymbol{n}} \quad \text { (notation of this book) } \tag{11.3} \end{equation*}(11.3)n=n (notation of this book) 
measures the separation between the geodesic n n nnn, regarded as the fiducial geodesic, and the typical nearby geodesic, n + Δ n n + Δ n n+Delta nn+\Delta nn+Δn (for small Δ n Δ n Delta n\Delta nΔn ), in the sense that
(11.4) ( Δ n ) n = { Δ n P n Δ n n } measures the change in { position any function } (11.4) ( Δ n ) n = Δ n P n Δ n n  measures the   change in   position   any   function  {:(11.4)(Delta n)n={[Delta n(delP)/(del n)],[Delta n(del)/(del n)]}{:[" measures the "],[" change in "]:}{[" position "],[" any "],[" function "]}:}(\Delta n) \boldsymbol{n}=\left\{\begin{array}{c} \Delta n \frac{\partial \mathscr{P}}{\partial n} \tag{11.4}\\ \Delta n \frac{\partial}{\partial n} \end{array}\right\} \begin{gathered} \text { measures the } \\ \text { change in } \end{gathered}\left\{\begin{array}{c} \text { position } \\ \text { any } \\ \text { function } \end{array}\right\}(11.4)(Δn)n={ΔnPnΔnn} measures the  change in { position  any  function }
brought about by transfer of attention from the one geodesic to the other at a fixed value of the affine parameter λ λ lambda\lambdaλ. This vector is represented by the arrow H Q H Q HQ\mathscr{H Q}HQ in the first diagram in Box 11.2.

Box 11.2 GEODESIC DEVIATION REPRESENTED AS AN ARROW

"Fiducial geodesic" n n nnn. Separation vector n Δ n = n Δ n = n Delta n=\boldsymbol{n} \Delta n=nΔn= R Q R Q RQ\mathscr{R} \mathscr{Q}RQ leads from point R R R\mathscr{R}R on it, to point Q Q Q\mathscr{Q}Q with same value of affine parameter λ λ lambda\lambdaλ (timelike quantity) on neighboring "test geodesic" n + Δ n n + Δ n n+Delta nn+\Delta nn+Δn.

Parallel transport of R 2 R 2 R2\mathscr{R 2}R2 by "Schild's ladder construction" (Box 10.2) to T O B T O B TOB\mathscr{T O} \mathscr{B}TOB and E a E a Ea\mathcal{E a}Ea. If the test geodesic n + Δ n n + Δ n n+Delta nn+\Delta nn+Δn had kept a constant separation from the fiducial geodesic n n nnn, its tracer point would have arrived at a a aaa at the value ( λ Δ λ ) ( λ Δ λ ) (lambda-Delta lambda)(\lambda-\Delta \lambda)(λΔλ) of the affine parameter, and at B B B\mathscr{B}B at ( λ + Δ λ ) ( λ + Δ λ ) (lambda+Delta lambda)(\lambda+\Delta \lambda)(λ+Δλ).
Actual location of tracer point of test geodesic at values of the timelike affine parameter ( λ Δ λ ) ( λ Δ λ ) (lambda-Delta lambda)(\lambda-\Delta \lambda)(λΔλ), λ λ lambda\lambdaλ, and ( λ + Δ λ ) ( λ + Δ λ ) (lambda+Delta lambda)(\lambda+\Delta \lambda)(λ+Δλ).
Confrontation between actual course of tracer point on test geodesic and "canonical course": course it would have had to take to keep constant separation from the tracer point moving along the fiducial geodesic.
Test geodesic same as before, except for uniform stretchout in scale of affine parameter. Any measure of departure of the actual course of geodesic from the canonical course ( a Q Z a Q Z aQZ\mathfrak{a Q} \mathcal{Z}aQZ ), to be useful, should be independent of this stretchout. Hence, take as measure of geodesic deviation, not the vector B R B R BR\mathscr{B} \mathscr{R}BR alone, nor the vector C P C P CP\mathscr{C P}CP, but the stretch-independent combination δ 2 = ( B R ) + δ 2 = ( B R ) + delta_(2)=(BR)+\boldsymbol{\delta}_{2}=(\mathscr{B R})+δ2=(BR)+ ( a P ) ( a P ) (aP)(a \mathscr{P})(aP). Here the sign of addition implies that the two vectors have been transported parallel to themselves, before addition, to a common point ( 2 in the diagram; R R R\mathscr{R}R in the differential calculus limit Δ n 0 , Δ λ 0 Δ n 0 , Δ λ 0 Delta n longrightarrow0,Delta lambda longrightarrow0\Delta n \longrightarrow 0, \Delta \lambda \longrightarrow 0Δn0,Δλ0 ).
Alternative courses that the test geodesic of D could have taken through Q Q Q\mathcal{Q}Q (families of geodesics characterized by different degrees of divergence from the left or convergence towards the right). Tilt changes values of Q P Q P QP\mathscr{Q P}QP (to A P A P ¯ A bar(P)\mathscr{A} \overline{\mathscr{P}}AP ) and B R B R BR\mathscr{B R}BR (to B R B R ¯ B bar(R)\mathscr{B} \overline{\mathscr{R}}BR ) individually, but not value of the sum δ 2 = ( B R ) δ 2 = ( B R ) delta_(2)=(BR)\boldsymbol{\delta}_{2}=(\mathscr{B} \mathscr{R})δ2=(BR) + ( a P ) + ( a P ) +(aP)+(a \mathscr{P})+(aP) ("lever principle").
Note that arrow B R B R BR\mathscr{B R}BR is of first order in Δ λ Δ λ Delta lambda\Delta \lambdaΔλ and of first order in Δ n Δ n Delta n\Delta nΔn; similarly for Q P Q P QP\mathscr{Q P}QP; hence the combination δ 2 δ 2 delta_(2)\boldsymbol{\delta}_{2}δ2 is of second order in Δ λ Δ λ Delta lambda\Delta \lambdaΔλ and first order in Δ n Δ n Delta n\Delta nΔn. Conclude that the arrow δ 2 / ( Δ λ ) 2 ( Δ n ) δ 2 / ( Δ λ ) 2 ( Δ n ) delta_(2)//(Delta lambda)^(2)(Delta n)\boldsymbol{\delta}_{2} /(\Delta \lambda)^{2}(\Delta n)δ2/(Δλ)2(Δn) is the desired measure of geodesic deviation in the sense that:
size of mesh (ultimately to go to zero) cancels out;
parameterization of test geodesic cancels out; slope of test geodesic cancels out.
Give this arrow the name "relative-acceleration vector"; and by examining it more closely (Box 11.3), discover the formula
δ 2 / ( Δ λ ) 2 ( Δ n ) = u u n δ 2 / ( Δ λ ) 2 ( Δ n ) = u u n delta_(2)//(Delta lambda)^(2)(Delta n)=grad_(u)grad_(u)n\boldsymbol{\delta}_{2} /(\Delta \lambda)^{2}(\Delta n)=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}δ2/(Δλ)2(Δn)=uun
for it.
Box 11.2 illustrates what it means to speak of geodesic deviation. One transports the separation n Δ n = H Q n Δ n = H Q n Delta n=HQn \Delta n=\mathscr{H} \mathscr{Q}nΔn=HQ parallel to itself along the fiducial geodesic. The tip of this vector traces out the canonical course that the nearby tracer point would have to pursue if it were to maintain constant separation from the fiducial tracer point. The actual course of the test geodesic deviates from this "canonical" course. The deviation, a vector ( A P A P AP\mathscr{A P}AP of Box 11.2), changes with the affine parameter ( A P P A P P APP\mathscr{A P P}APP at a a a\mathscr{a}a, 0 at Q , B R Q , B R Q,BR\mathscr{Q}, \mathscr{B} \mathscr{R}Q,BR at B ) B ) B)\mathscr{B})B). The first derivative of this vector with respect to the affine parameter is sensitive to the scale of parameterization along the test geodesic, and to its slope (Box 11.2, F). Not so the second derivative. It depends only on the tangent vector u u u\boldsymbol{u}u of the fiducial geodesic, and on the separation vector n Δ n n Δ n n Delta n\boldsymbol{n} \Delta nnΔn. Divide this second derivative of the deviation by Δ n Δ n Delta n\Delta nΔn and give it a name: the "relative-acceleration vector". Discover (Box 11.3) a simple formula for it
(11.5) (relative-acceleration vector) = u u n (11.5)  (relative-acceleration vector)  = u u n {:(11.5)" (relative-acceleration vector) "=grad_(u)grad_(u)n:}\begin{equation*} \text { (relative-acceleration vector) }=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n} \tag{11.5} \end{equation*}(11.5) (relative-acceleration vector) =uun

§11.3. TIDAL GRAVITATIONAL FORCES AND RIEMANN CURVATURE TENSOR

With "relative acceleration" now defined, turn to the "tidal gravitational force" (i.e., "spacetime curvature") that produces it. Use a Newtonian analysis of tidal forces

Box 11.3 GEODESIC DEVIATION: ARROW CORRELATED WITH SECOND COVARIANT DERIVATIVE

The arrow δ 2 δ 2 delta_(2)\boldsymbol{\delta}_{2}δ2 in Box 11.2 measures, not the rate of change of the separation of the test geodesic n + Δ n n + Δ n n+Delta nn+\Delta nn+Δn from the "canonical course" Q Q Q Q Q Q QQQ\mathscr{Q Q Q}QQQ as baseline, but the second derivative:
( first derivative at λ + 1 2 Δ λ ) = u n = R R B Δ λ Δ n = B R Δ λ Δ n ( first derivative at λ 1 2 Δ λ ) = u n = E Q E P Δ λ Δ n = Q P Δ λ Δ n  first derivative at  λ + 1 2 Δ λ = u n = R R B Δ λ Δ n = B R Δ λ Δ n  first derivative at  λ 1 2 Δ λ = u n = E Q E P Δ λ Δ n = Q P Δ λ Δ n {:[(" first derivative at "lambda+(1)/(2)Delta lambda)=grad_(u)n=(R-R-B)/(Delta lambda Delta n)=(BR)/(Delta lambda Delta n)],[(" first derivative at "lambda-(1)/(2)Delta lambda)=grad_(u)n=(EQ-EP)/(Delta lambda Delta n)=(-QP)/(Delta lambda Delta n)]:}\begin{aligned} & \left(\text { first derivative at } \lambda+\frac{1}{2} \Delta \lambda\right)=\boldsymbol{\nabla}_{u} \boldsymbol{n}=\frac{\mathscr{R}-\mathscr{R}-\mathscr{B}}{\Delta \lambda \Delta n}=\frac{\mathscr{B} \mathscr{R}}{\Delta \lambda \Delta n} \\ & \left(\text { first derivative at } \lambda-\frac{1}{2} \Delta \lambda\right)=\boldsymbol{\nabla}_{u} \boldsymbol{n}=\frac{\mathscr{E Q}-\mathscr{E} \mathscr{P}}{\Delta \lambda \Delta n}=\frac{-\mathscr{Q} \mathscr{P}}{\Delta \lambda \Delta n} \end{aligned}( first derivative at λ+12Δλ)=un=RRBΔλΔn=BRΔλΔn( first derivative at λ12Δλ)=un=EQEPΔλΔn=QPΔλΔn
Transpose to common location λ λ lambda\lambdaλ, take difference, and divide it by Δ λ Δ λ Delta lambda\Delta \lambdaΔλ to obtain the second covariant derivative with respect to the vector u u u\boldsymbol{u}u; thus
u u n = ( u n ) λ + 1 2 Δ λ ( u n ) λ 1 2 Δ λ Δ λ = ( B R + A P ) vectors transported to common location ( Δ λ ) 2 Δ n = δ 2 ( Δ λ ) 2 Δ n = "relative acceleration vector" for neighboring geodesics. u u n = u n λ + 1 2 Δ λ u n λ 1 2 Δ λ Δ λ = ( B R + A P )  vectors transported to  common location  ( Δ λ ) 2 Δ n = δ 2 ( Δ λ ) 2 Δ n =  "relative acceleration vector" for neighboring geodesics.  {:[grad_(u)grad_(u)n=((grad_(u)n)_(lambda+(1)/(2)Delta lambda)-(grad_(u)n)_(lambda-(1)/(2))Delta lambda)/(Delta lambda)],[=((BR+AP)_({:" vectors transported to ":})^("common location "))/((Delta lambda)^(2)Delta n)=(delta_(2))/((Delta lambda)^(2)Delta n)],[=" "relative acceleration vector" for neighboring geodesics. "]:}\begin{aligned} \boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} \boldsymbol{n} & =\frac{\left(\boldsymbol{\nabla}_{u} \boldsymbol{n}\right)_{\lambda+\frac{1}{2} \Delta \lambda}-\left(\boldsymbol{\nabla}_{u} \boldsymbol{n}\right)_{\lambda-\frac{1}{2}} \Delta \lambda}{\Delta \lambda} \\ & =\frac{(\mathscr{B R}+\mathscr{A P})_{\begin{array}{l} \text { vectors transported to } \end{array}}^{\text {common location }}}{(\Delta \lambda)^{2} \Delta n}=\frac{\boldsymbol{\delta}_{2}}{(\Delta \lambda)^{2} \Delta n} \\ & =\text { "relative acceleration vector" for neighboring geodesics. } \end{aligned}uun=(un)λ+12Δλ(un)λ12ΔλΔλ=(BR+AP) vectors transported to common location (Δλ)2Δn=δ2(Δλ)2Δn= "relative acceleration vector" for neighboring geodesics. 
(left half of Box 11.4) to motivate the geometric analysis (right half of same box). Thereby arrive at the remarkable equation
This equation is remarkable, because at first sight it seems crazy. The term [ n , u ] u n , u u [grad_(n),grad_(u)]u\left[\boldsymbol{\nabla}_{n}, \boldsymbol{\nabla}_{u}\right] \boldsymbol{u}[n,u]u involves second derivatives of u u u\boldsymbol{u}u, and a first derivative of n n grad_(n)\boldsymbol{\nabla}_{\boldsymbol{n}}n :
(11.7) [ n , u ] u n u u u n u (11.7) n , u u n u u u n u {:(11.7)[grad_(n),grad_(u)]u-=grad_(n)grad_(u)u-grad_(u)grad_(n)u:}\begin{equation*} \left[\nabla_{n}, \nabla_{u}\right] u \equiv \nabla_{n} \nabla_{u} u-\nabla_{u} \nabla_{n} u \tag{11.7} \end{equation*}(11.7)[n,u]unuuunu
It thus must depend on how u u u\boldsymbol{u}u and n n n\boldsymbol{n}n vary from point to point. But the relative acceleration it produces, u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{u}} \nabla_{\boldsymbol{u}} \boldsymbol{n}uun, is known to depend only on the values of u u u\boldsymbol{u}u and n n n\boldsymbol{n}n at the fiducial point, not on how u u u\boldsymbol{u}u and n n n\boldsymbol{n}n vary (see Box 11.2, F). How is this possible?
Somehow all derivatives must drop out of the tidal-force quantity [ n , u ] u n , u u [grad_(n),grad_(u)]u\left[\boldsymbol{\nabla}_{\boldsymbol{n}}, \boldsymbol{\nabla}_{\boldsymbol{u}}\right] \boldsymbol{u}[n,u]u. One must be able to regard [ , . , ] , . , [grad_(dots),.,grad_(dots)]dots\left[\boldsymbol{\nabla}_{\ldots}, ., \boldsymbol{\nabla}_{\ldots}\right] \ldots[,.,] as a purely local, algebraic machine with three slots, whose output is a vector. If it is purely local and not differential, then it is even linear (as one sees from the additivity properties of grad\mathbf{\nabla} ), so it must be a tensor. Give this tensor the name Riemann, and give it a fourth slot for inputting a 1-form:
Riemann ( , C , A , B ) [ A , B ] C Riemann ( σ , C , A , B ) σ , [ A , B ] C .  Riemann  ( , C , A , B ) A , B C  Riemann  ( σ , C , A , B ) σ , A , B C . {:[" Riemann "(dots","C","A","B)-=[grad_(A),grad_(B)]C],[" Riemann "(sigma","C","A","B)-=(:sigma,[grad_(A),grad_(B)]C:).]:}\begin{aligned} \text { Riemann }(\ldots, C, A, B) & \equiv\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{B}\right] C \\ \text { Riemann }(\sigma, C, A, B) & \equiv\left\langle\sigma,\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{B}\right] C\right\rangle . \end{aligned} Riemann (,C,A,B)[A,B]C Riemann (σ,C,A,B)σ,[A,B]C.
This is only a tentative definition of Riemann. Before accepting it, one should verify that it is, indeed, a tensor. Does it really depend on only the values of A A A\boldsymbol{A}A, B , C B , C B,C\boldsymbol{B}, \boldsymbol{C}B,C at the point of evaluation, and not on how they are changing there? The answer (derived in Box 11.5) is "almost." It fails the test, but with a slight modification it will pass. The modification is to replace the commutator [ A , B ] A , B [grad_(A),grad_(B)]\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right][A,B] by the "curvature operator"
(11.8) R ( A , B ) [ A , B ] [ A , B ] , (11.8) R ( A , B ) A , B [ A , B ] , {:(11.8)R(A","B)-=[grad_(A),grad_(B)]-grad_([A,B])",":}\begin{equation*} \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \equiv\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{B}\right]-\boldsymbol{\nabla}_{[A, B]}, \tag{11.8} \end{equation*}(11.8)R(A,B)[A,B][A,B],
where [ A , B ] [ A , B ] grad_([A,B])\boldsymbol{\nabla}_{[\boldsymbol{A}, \boldsymbol{B}]}[A,B] is the derivative along the vector [ A , B ] [ A , B ] [A,B][\boldsymbol{A}, \boldsymbol{B}][A,B] (commutator of A A A\boldsymbol{A}A and B B B\boldsymbol{B}B ). ( ( A , B ) [ A , B ] ( A , B ) A , B (ℜ(A,B)-=[grad_(A),grad_(B)]:}\left(\Re(\boldsymbol{A}, \boldsymbol{B}) \equiv\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right]\right.((A,B)[A,B] for the fields A = n A = n A=n\boldsymbol{A}=\boldsymbol{n}A=n and B = u B = u B=u\boldsymbol{B}=\boldsymbol{u}B=u of the geodesic-deviation problem, because [ n , u ] = 0 [ n , u ] = 0 [n,u]=0[\boldsymbol{n}, \boldsymbol{u}]=0[n,u]=0.) Then the modified and acceptable definition of the Riemann curvature tensor is
Riemann ( , C , A , B ) R ( A , B ) C (11.9) Riemann ( σ , C , A , B ) σ , R ( A , B ) C  Riemann  ( , C , A , B ) R ( A , B ) C (11.9)  Riemann  ( σ , C , A , B ) σ , R ( A , B ) C {:[" Riemann "(dots","C","A","B)-=R(A","B)C],[(11.9)" Riemann "(sigma","C","A","B)-=(:sigma","R(A","B)C:)]:}\begin{align*} \text { Riemann }(\ldots, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B}) & \equiv \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C} \\ \text { Riemann }(\boldsymbol{\sigma}, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B}) & \equiv\langle\boldsymbol{\sigma}, \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}\rangle \tag{11.9} \end{align*} Riemann (,C,A,B)R(A,B)C(11.9) Riemann (σ,C,A,B)σ,R(A,B)C
To define Riemann thus, and to verify its tensorial character (exercise 11.2), does not by any means teach one what curvature is all about. To understand curvature, one must scrutinize Riemann from all viewpoints. That is the task of the rest of this chapter.
Curvature operator defined
Riemann curvature tensor defined
Box 11.4 RELATIVE ACCELERATION OF TEST PARTICLESGEOMETRIC ANALYSIS PATTERNED ON NEWTONIAN ANALYSIS

Newtonian Analysis

  1. Consider a family of test-particle trajectories x j ( t , n ) x j ( t , n ) x^(j)(t,n)x^{j}(t, n)xj(t,n) in ordinary, three-dimensional space: " t t ttt " is time measured by particle's clock, or any clock; " n n nnn " is "selector parameter."
  2. Equation of motion for each trajectory:
( 2 x j t 2 ) n + Φ x j = 0 2 x j t 2 n + Φ x j = 0 ((del^(2)x^(j))/(delt^(2)))_(n)+(del Phi)/(delx^(j))=0\left(\frac{\partial^{2} x^{j}}{\partial t^{2}}\right)_{n}+\frac{\partial \Phi}{\partial x^{j}}=0(2xjt2)n+Φxj=0
where Φ Φ Phi\PhiΦ is Newtonian potential.
3. Take difference between equations of motion for neighboring trajectories, n n nnn and n + Δ n n + Δ n n+Delta nn+\Delta nn+Δn, and take limit as Δ n 0 Δ n 0 Delta n longrightarrow0\Delta n \longrightarrow 0Δn0-i.e., take derivative
( n ) t [ ( 2 x j t 2 ) n + Φ x j ] = 0 . n t 2 x j t 2 n + Φ x j = 0 . ((del)/(del n))_(t)[((del^(2)x^(j))/(delt^(2)))_(n)+(del Phi)/(delx^(j))]=0.\left(\frac{\partial}{\partial n}\right)_{t}\left[\left(\frac{\partial^{2} x^{j}}{\partial t^{2}}\right)_{n}+\frac{\partial \Phi}{\partial x^{j}}\right]=0 .(n)t[(2xjt2)n+Φxj]=0.
  1. When / n / n del//del n\partial / \partial n/n acts on second term, rewrite it as
( n ) t = ( x k n ) t x k = n k x k n t = x k n t x k = n k x k ((del)/(del n))_(t)=((delx^(k))/(del n))_(t)(del)/(delx^(k))=n^(k)(del)/(delx^(k))\left(\frac{\partial}{\partial n}\right)_{t}=\left(\frac{\partial x^{k}}{\partial n}\right)_{t} \frac{\partial}{\partial x^{k}}=n^{k} \frac{\partial}{\partial x^{k}}(n)t=(xkn)txk=nkxk
Thereby obtain
( n ) t ( t ) n ( x j t ) n + 2 Φ x j x k n k = 0 . n t t n x j t n + 2 Φ x j x k n k = 0 . ((del)/(del n))_(t)((del)/(del t))_(n)((delx^(j))/(del t))_(n)+(del^(2)Phi)/(delx^(j)delx^(k))n^(k)=0.\left(\frac{\partial}{\partial n}\right)_{t}\left(\frac{\partial}{\partial t}\right)_{n}\left(\frac{\partial x^{j}}{\partial t}\right)_{n}+\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} n^{k}=0 .(n)t(t)n(xjt)n+2Φxjxknk=0.

Geometric Analysis

  1. Consider a family of test-particle trajectories (geodesics), P ( λ , n ) P ( λ , n ) P(lambda,n)\mathscr{P}(\lambda, n)P(λ,n), in spacetime: " λ λ lambda\lambdaλ " is affineparameter, i.e., time measured by particle's clock; " n n nnn " is "selector parameter."
  2. Geodesic equation for each trajectory:
u u = 0 . u u = 0 . grad_(u)u=0.\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0 .uu=0.
[Looks like first-order equation; is actually second-order because the " u u u\boldsymbol{u}u " being differentiated is itself a derivative, u = ( P / λ ) n u = ( P / λ ) n u=(delP//del lambda)_(n)\boldsymbol{u}=(\partial \mathscr{P} / \partial \lambda)_{n}u=(P/λ)n.]
3. Take difference between geodesic equations for neighboring geodesics n n nnn and n + Δ n n + Δ n n+Delta nn+\Delta nn+Δn, and take limit as Δ n 0 Δ n 0 Delta n longrightarrow0\Delta n \longrightarrow 0Δn0-i.e., take covariant derivative
n [ u u ] = 0 n u u = 0 grad_(n)[grad_(u)u]=0\boldsymbol{\nabla}_{n}\left[\boldsymbol{\nabla}_{u} \boldsymbol{u}\right]=0n[uu]=0
  1. There is no second term, so leave equation in form
n [ u u ] = 0 n u u = 0 grad_(n)[grad_(u)u]=0\boldsymbol{\nabla}_{n}\left[\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}\right]=0n[uu]=0
  1. To obtain equation for relative acceleration, move ( / n ) t ( / n ) t (del//del n)_(t)(\partial / \partial n)_{t}(/n)t through both of the ( / t ) n ( / t ) n (del//del t)_(n)(\partial / \partial t)_{n}(/t)n terms (permissible because partial derivatives commute!):
( t ) n ( t ) n ( x j n ) t + 2 Φ x j x k n k = 0 . t n t n x j n t + 2 Φ x j x k n k = 0 . ((del)/(del t))_(n)((del)/(del t))_(n)((delx^(j))/(del n))_(t)+(del^(2)Phi)/(delx^(j)delx^(k))n^(k)=0.\left(\frac{\partial}{\partial t}\right)_{n}\left(\frac{\partial}{\partial t}\right)_{n}\left(\frac{\partial x^{j}}{\partial n}\right)_{t}+\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} n^{k}=0 .(t)n(t)n(xjn)t+2Φxjxknk=0.
This is equivalent to

5. To obtain equation for relative acceleration, u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}uun, move n n grad_(n)\boldsymbol{\nabla}_{\boldsymbol{n}}n through u u grad_(u)\boldsymbol{\nabla}_{\boldsymbol{u}}u and through the / λ / λ del//del lambda\partial / \partial \lambda/λ of u = P / λ : u = P / λ : u=delP//del lambda:\boldsymbol{u}=\partial \mathscr{P} / \partial \lambda:u=P/λ:
a. First step: In n u u = 0 In n u u = 0 Ingrad_(n)grad_(u)u=0\operatorname{In} \boldsymbol{\nabla}_{\boldsymbol{n}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0Innuu=0, move n n grad_(n)\boldsymbol{\nabla}_{\boldsymbol{n}}n through u u grad_(u)\boldsymbol{\nabla}_{\boldsymbol{u}}u. The result:

commutator; must be included as protection against possibility that u n n u u n n u grad_(u)grad_(n)!=grad_(n)grad_(u)\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{n} \neq \boldsymbol{\nabla}_{n} \boldsymbol{\nabla}_{u}unnu.
b. Second step: Move n n grad_(n)\boldsymbol{\nabla}_{\boldsymbol{n}}n through / λ / λ del//del lambda\partial / \partial \lambda/λ of u = P / λ u = P / λ u=delP//del lambda\boldsymbol{u}=\partial \mathscr{P} / \partial \lambdau=P/λ; i.e., write
Why? Because symmetry of covariant derivative says n u u n = [ n , u ] n u u n = [ n , u ] grad_(n)u-grad_(u)n=[n,u]\boldsymbol{\nabla}_{\boldsymbol{n}} \boldsymbol{u}-\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}=[\boldsymbol{n}, \boldsymbol{u}]nuun=[n,u]
= [ n , λ ] = 2 n λ 2 λ n = 0 . = n , λ = 2 n λ 2 λ n = 0 . =[(del)/(del n),(del)/(del lambda)]=(del^(2))/(del n del lambda)-(del^(2))/(del lambda del n)=0.=\left[\frac{\partial}{\partial n}, \frac{\partial}{\partial \lambda}\right]=\frac{\partial^{2}}{\partial n \partial \lambda}-\frac{\partial^{2}}{\partial \lambda \partial n}=0 .=[n,λ]=2nλ2λn=0.
c. Result:

Box 11.5 RIEMANN CURVATURE TENSOR

A. Definition of Riemann Motivated by Tidal Gravitational Forces:

  1. Tidal forces (spacetime curvature) produce relative acceleration of test particles (geodesics) given by
(1) u u n + [ n , u ] u = 0 . (1) u u n + n , u u = 0 . {:(1)grad_(u)grad_(u)n+[grad_(n),grad_(u)]u=0.:}\begin{equation*} \boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} n+\left[\boldsymbol{\nabla}_{n}, \boldsymbol{\nabla}_{u}\right] \boldsymbol{u}=0 . \tag{1} \end{equation*}(1)uun+[n,u]u=0.

Box 11.5 (continued)

  1. This motivates the definition
(2) Riemann ( , C , A , B ) = [ A , B ] C . (2)  Riemann  ( , C , A , B ) = A , B C . {:(2)" Riemann "(dots","C","A","B)=[grad_(A),grad_(B)]C.:}\begin{equation*} \text { Riemann }(\ldots, C, A, B)=\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{B}\right] C . \tag{2} \end{equation*}(2) Riemann (,C,A,B)=[A,B]C.
B. Failure of this Definition
  1. Definition acceptable only if Riemann ( , C , A , B ) ( , C , A , B ) (dots,C,A,B)(\ldots, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B})(,C,A,B) is a linear machine, independent of how A , B , C A , B , C A,B,C\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}A,B,C vary from point to point.
  2. Check, in part: change variations of C C C\boldsymbol{C}C, but not C C C\boldsymbol{C}C itself, at event P 0 P 0 P_(0)\mathscr{P}_{0}P0 :
C NEW ( P ) = f ( P ) C oLD ( P ) . Larbitrary function except f ( P 0 ) = 1 ] C NEW ( P ) = f ( P ) C oLD  ( P ) . Larbitrary function except  f P 0 = 1 {:[C_(NEW)(P)=f(P)C_("oLD ")(P).],[{:_("Larbitrary function except ")f(P_(0))=1]]:}\begin{aligned} \boldsymbol{C}_{\mathrm{NEW}}(\mathscr{P})= & f(\mathscr{P}) \boldsymbol{C}_{\text {oLD }}(\mathscr{P}) . \\ & \text { } \left._{\text {Larbitrary function except }} f\left(\mathscr{P}_{0}\right)=1\right] \end{aligned}CNEW(P)=f(P)CoLD (P). Larbitrary function except f(P0)=1]
  1. Does this change [ A , B ] C A , B C [grad_(A),grad_(B)]C\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right] \boldsymbol{C}[A,B]C ? Yes! Exercise 11.1 shows
{ [ A , B ] C NEW } at P 0 { [ A , B ] C OLD } P 0 = C OLD [ A , B ] f . A , B C NEW at  P 0 A , B C OLD  P 0 = C OLD  [ A , B ] f . {[grad_(A),grad_(B)]C_(NEW)}_("at "P_(0))-{[grad_(A),grad_(B)]C_("OLD ")}_(P_(0))=C_("OLD ")grad_([A,B])f.\left\{\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{B}\right] \boldsymbol{C}_{\mathrm{NEW}}\right\}_{\text {at } \mathscr{P}_{0}}-\left\{\left[\boldsymbol{\nabla}_{A}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right] \boldsymbol{C}_{\text {OLD }}\right\}_{\mathscr{P}_{0}}=\boldsymbol{C}_{\text {OLD }} \boldsymbol{\nabla}_{[A, B]} f .{[A,B]CNEW}at P0{[A,B]COLD }P0=COLD [A,B]f.

C. Modified Definition of Riemann:

  1. The term causing trouble, C oLD [ A , B ] f C oLD  [ A , B ] f C_("oLD ")grad_([A,B])f\boldsymbol{C}_{\text {oLD }} \boldsymbol{\nabla}_{[\boldsymbol{A}, \boldsymbol{B}]} fCoLD [A,B]f, can be disposed of by subtracting a "correction term" resembling it from Riemann-i.e., by redefining
Riemann (3) ( , C , A , B ) R ( A , B ) C (4) R ( A , B ) [ A , B ] [ A , B ]  Riemann  (3) ( , C , A , B ) R ( A , B ) C (4) R ( A , B ) A , B [ A , B ] " Riemann "{:[(3)(dots","C","A","B)-=R(A","B)C],[(4)R(A","B)-=[grad_(A),grad_(B)]-grad_([A,B])]:}\text { Riemann } \begin{align*} (\ldots, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B}) & \equiv \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C} \tag{3}\\ \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) & \equiv\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{A}, \boldsymbol{B}]} \tag{4} \end{align*} Riemann (3)(,C,A,B)R(A,B)C(4)R(A,B)[A,B][A,B]
  1. The above calculation then gives a result independent of the "modifying function" f f fff :
{ ( A , B ) C NEW } at Φ 0 = { R ( A , B ) C OLD } at Φ 0 . ( A , B ) C NEW at Φ 0 = R ( A , B ) C OLD at Φ 0 . {ℜ(A,B)C_(NEW)}_(atPhi_(0))={R(A,B)C_(OLD)}_(atPhi_(0)).\left\{\Re(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}_{\mathrm{NEW}}\right\}_{\mathrm{at} \Phi_{0}}=\left\{\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}_{\mathrm{OLD}}\right\}_{\mathrm{at} \Phi_{0}} .{(A,B)CNEW}atΦ0={R(A,B)COLD}atΦ0.

D. Is Modified Definition Compatible with Equation for Tidal Gravitational Forces?

  1. One would like to write u u n + R i e m a n n ( , u , n , u ) = 0 u u n + R i e m a n n ( , u , n , u ) = 0 grad_(u)grad_(u)n+Riemann(dots,u,n,u)=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}+\boldsymbol{\operatorname { R i e m a n n }}(\ldots, \boldsymbol{u}, \boldsymbol{n}, \boldsymbol{u})=0uun+Riemann(,u,n,u)=0.
  2. This works just as well for modified definition of Riemann as for original definition, because
R ( n , u ) = [ n , u ] [ n , u ] = [ n , u ] = 0 because n = ( / n ) λ and u = ( / λ ) n commute ] R ( n , u ) = n , u [ n , u ] = n , u = 0  because  n = ( / n ) λ  and  u = ( / λ ) n  commute  {:[R(n","u)=[grad_(n),grad_(u)]-grad_([n,u])=[grad_(n),grad_(u)]],[ uarr],[=0" because "n=(del//del n)_(lambda)" and "],[{:u=(del//del lambda)_(n)" commute "]]:}\begin{aligned} & \mathscr{R}(\boldsymbol{n}, \boldsymbol{u})=\left[\boldsymbol{\nabla}_{\boldsymbol{n}}, \boldsymbol{\nabla}_{\boldsymbol{u}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{n}, \boldsymbol{u}]}=\left[\mathbf{\nabla}_{\boldsymbol{n}}, \boldsymbol{\nabla}_{\boldsymbol{u}}\right] \\ & \uparrow \\ &=0 \text { because } \boldsymbol{n}=(\partial / \partial n)_{\lambda} \text { and } \\ &\left.\boldsymbol{u}=(\partial / \partial \lambda)_{n} \text { commute }\right] \end{aligned}R(n,u)=[n,u][n,u]=[n,u]=0 because n=(/n)λ and u=(/λ)n commute ]
Geodesic deviation and tidal forces cannot tell the difference between R ( n , u ) R ( n , u ) R(n,u)\mathscr{R}(\boldsymbol{n}, \boldsymbol{u})R(n,u) and [ n , u ] n , u [grad_(n),grad_(u)]\left[\boldsymbol{\nabla}_{\boldsymbol{n}}, \boldsymbol{\nabla}_{u}\right][n,u], nor consequently between old and new definitions of Riemann.

E. Is Modified Definition Acceptable?

I.e., is Riemann ( , C , A , B ) R ( A , B ) C ( , C , A , B ) R ( A , B ) C (dots,C,A,B)-=R(A,B)C(\ldots, \boldsymbol{C}, \boldsymbol{A}, \boldsymbol{B}) \equiv \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}(,C,A,B)R(A,B)C a linear machine with output independent of how A , B , C A , B , C A,B,C\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}A,B,C vary near point of evaluation? YES! (See exercise 11.2.)
Take stock, first, of what one knows already about the Riemann curvature tensor. (1) Riemann is a tensor; despite the appearance of grad\boldsymbol{\nabla} in its definition (11.9), no derivatives actually act on the input vectors A , B A , B A,B\boldsymbol{A}, \boldsymbol{B}A,B, and C C C\boldsymbol{C}C. (2) Riemann is a ( 1 3 ) 1 3 ((1)/(3))\left(\frac{1}{3}\right)(13) tensor; its first slot accepts a 1 -form; the others, vectors. (3) Riemann is determined entirely by grad\boldsymbol{\nabla}, or equivalently by the geodesics of spacetime, or equivalently by spacetime's parallel transport law; nothing but grad\boldsymbol{\nabla} and the input vectors and 1-form are required to fix Riemann's output. (4) Riemann produces the tidal gravitational forces that pry geodesics (test-particle trajectories) apart or push them together; i.e., it characterizes the "curvature of spacetime":
(11.10) u u n + Riemann ( , u , n , u ) = 0 (11.10) u u n +  Riemann  ( , u , n , u ) = 0 {:(11.10)grad_(u)grad_(u)n+" Riemann "(dots","u","n","u)=0:}\begin{equation*} \nabla_{u} \nabla_{u} n+\text { Riemann }(\ldots, u, n, u)=0 \tag{11.10} \end{equation*}(11.10)uun+ Riemann (,u,n,u)=0
(This "equation of geodesic deviation" follows from equations 11.6, 11.8, and 11.9, and the relation [ n , u ] = 0 [ n , u ] = 0 [n,u]=0[\boldsymbol{n}, \boldsymbol{u}]=0[n,u]=0.)
All these facets of Riemann are pictorial (e.g., geodesic deviation; see Boxes 11.2 and 11.3) or abstract (e.g., equations 11.8 and 11.9 for Riemann in terms of grad\boldsymbol{\nabla} ). Riemann's component facet,
(11.11) R β γ δ α Riemann ( ω α , e β , e γ , e δ ) ω α , R ( e γ , e δ ) e β (11.11) R β γ δ α Riemann ω α , e β , e γ , e δ ω α , R e γ , e δ e β {:(11.11)R_(beta gamma delta)^(alpha)-=Riemann(omega^(alpha),e_(beta),e_(gamma),e_(delta))-=(:omega^(alpha),R(e_(gamma),e_(delta))e_(beta):):}\begin{equation*} R_{\beta \gamma \delta}^{\alpha} \equiv \operatorname{Riemann}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \equiv\left\langle\boldsymbol{\omega}^{\alpha}, \mathscr{R}\left(\boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \boldsymbol{e}_{\beta}\right\rangle \tag{11.11} \end{equation*}(11.11)RβγδαRiemann(ωα,eβ,eγ,eδ)ωα,R(eγ,eδ)eβ
is related to the component facet of grad\boldsymbol{\nabla} by the following equation, valid in any coordinate basis { e α } = { / x α } e α = / x α {e_(alpha)}={del//delx^(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}=\left\{\partial / \partial x^{\alpha}\right\}{eα}={/xα} :
(11.12) R β γ δ α = Γ β δ α x γ Γ β γ α x δ + Γ α μ γ Γ β δ μ Γ α μ δ Γ β γ μ (11.12) R β γ δ α = Γ β δ α x γ Γ β γ α x δ + Γ α μ γ Γ β δ μ Γ α μ δ Γ β γ μ {:(11.12)R_(beta gamma delta)^(alpha)=(delGamma_(beta delta)^(alpha))/(delx^(gamma))-(delGamma_(beta gamma)^(alpha))/(delx^(delta))+Gamma^(alpha)_(mu gamma)Gamma_(beta delta)^(mu)-Gamma^(alpha)_(mu delta)Gamma_(beta gamma)^(mu):}\begin{equation*} R_{\beta \gamma \delta}^{\alpha}=\frac{\partial \Gamma_{\beta \delta}^{\alpha}}{\partial x^{\gamma}}-\frac{\partial \Gamma_{\beta \gamma}^{\alpha}}{\partial x^{\delta}}+\Gamma^{\alpha}{ }_{\mu \gamma} \Gamma_{\beta \delta}^{\mu}-\Gamma^{\alpha}{ }_{\mu \delta} \Gamma_{\beta \gamma}^{\mu} \tag{11.12} \end{equation*}(11.12)Rβγδα=ΓβδαxγΓβγαxδ+ΓαμγΓβδμΓαμδΓβγμ
(See exercise 11.3 for derivation, and exercise 11.4 for the extension to noncoordinate bases.) These components of Riemann, with no sign of any derivative operator anywhere, may leave one with a better feeling in one's stomach than the definition (11.8) with its nondifferentiating derivatives!
Tide-producing gravitational forces expressed in terms of Riemann
Components of Riemann expressed in terms of connection coefficients

EXERCISES

Exercise 11.1. [ A , B ] C A , B C [grad_(A),grad_(B)]C\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right] \boldsymbol{C}[A,B]C DEPENDS ON DERIVATIVES OF C C C\boldsymbol{C}C

(Based on Box 11.5.) Let C NEW C NEW C_(NEW)\boldsymbol{C}_{\mathrm{NEW}}CNEW and C oLD C oLD  C_("oLD ")\boldsymbol{C}_{\text {oLD }}CoLD  be vector fields related by
C NEW ( P ) = f ( P ) C OLD ( P ) . arbitrary function, except f ( P 0 ) = 1 ] C NEW ( P ) = f ( P ) C OLD ( P ) . arbitrary function, except  f P 0 = 1 {:[C_(NEW)(P)=f(P)C_(OLD)(P).],[⨆_("arbitrary function, except "{:f(P_(0))=1])]:}\begin{aligned} \boldsymbol{C}_{\mathrm{NEW}}(\mathscr{P})= & f(\mathscr{P}) \boldsymbol{C}_{\mathrm{OLD}}(\mathscr{P}) . \\ & \bigsqcup_{\text {arbitrary function, except } \left.f\left(\mathscr{P}_{0}\right)=1\right]} \end{aligned}CNEW(P)=f(P)COLD(P).arbitrary function, except f(P0)=1]
Show that
{ [ A , B ] c NEWW } at Q 0 { [ A , B ] C OLD } at Q 0 = c OLD [ A , B ] f . A , B c NEWW at Q 0 A , B C OLD  at Q 0 = c OLD  [ A , B ] f . {[grad_(A),grad_(B)]c_(NEWW)}_(atQ_(0))-{[grad_(A),grad_(B)]C_("OLD ")}_(atQ_(0))=c_("OLD ")grad_([A,B])f.\left\{\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right] \boldsymbol{c}_{\mathrm{NEWW}}\right\}_{\mathrm{at} \mathscr{Q}_{0}}-\left\{\left[\boldsymbol{\nabla}_{\boldsymbol{A}}, \boldsymbol{\nabla}_{\boldsymbol{B}}\right] \boldsymbol{C}_{\text {OLD }}\right\}_{\mathrm{at} \mathscr{Q}_{0}}=\boldsymbol{c}_{\text {OLD }} \boldsymbol{\nabla}_{[A, B]} f .{[A,B]cNEWW}atQ0{[A,B]COLD }atQ0=cOLD [A,B]f.

Exercise 11.2. PROOF THAT Riemann IS A TENSOR

Show from its definition ( 11.8 , 11.9 ) ( 11.8 , 11.9 ) (11.8,11.9)(11.8,11.9)(11.8,11.9) that Riemann is a tensor. Hint: Use the following procedure.
(a) If f ( F ) f ( F ) f(F)f(\mathscr{F})f(F) is an arbitrary function, show that
R ( A , B ) f C = f R ( A , B ) C . R ( A , B ) f C = f R ( A , B ) C . R(A,B)fC=fR(A,B)C.\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) f \boldsymbol{C}=f \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C} .R(A,B)fC=fR(A,B)C.
(b) Similarly show that
M ( f A , B ) C = f M ( A , B ) C and R ( A , f B ) C = f ( A , B ) C . M ( f A , B ) C = f M ( A , B ) C  and  R ( A , f B ) C = f ( A , B ) C . M(fA,B)C=fM(A,B)C quad" and "quadR(A,fB)C=f(A,B)C.\mathscr{M}(f \boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}=f \mathscr{M}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C} \quad \text { and } \quad \mathscr{R}(\boldsymbol{A}, f \boldsymbol{B}) \boldsymbol{C}=f \mathscr{}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C} .M(fA,B)C=fM(A,B)C and R(A,fB)C=f(A,B)C.
(c) Show that R ( A , B ) C R ( A , B ) C R(A,B)C\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}R(A,B)C is linear; i.e.,
R ( A + a , B ) C = R ( A , B ) C + R ( a , B ) C R ( A , B + b ) C = R ( A , B ) C + R ( A , b ) C R , A , B ) ( C + c ) = R ( A , B ) C + R ( A , B ) c . R ( A + a , B ) C = R ( A , B ) C + R ( a , B ) C R ( A , B + b ) C = R ( A , B ) C + R ( A , b ) C R , A , B ) ( C + c ) = R ( A , B ) C + R ( A , B ) c . {:[R(A+a","B)C=R(A","B)C+R(a","B)C],[R(A","B+b)C=R(A","B)C+R(A","b)C],[R","A","B)(C+c)=R(A","B)C+R(A","B)c.]:}\begin{aligned} \mathscr{R}(\boldsymbol{A}+\boldsymbol{a}, \boldsymbol{B}) \boldsymbol{C} & =\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}+\mathscr{R}(\boldsymbol{a}, \boldsymbol{B}) \boldsymbol{C} \\ \mathscr{R}(\boldsymbol{A}, \boldsymbol{B}+\boldsymbol{b}) \boldsymbol{C} & =\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}+\mathscr{R}(\boldsymbol{A}, \boldsymbol{b}) \boldsymbol{C} \\ \mathscr{R}, \boldsymbol{A}, \boldsymbol{B})(\boldsymbol{C}+\boldsymbol{c}) & =\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}+\mathscr{R}(\mathbf{A}, \boldsymbol{B}) \boldsymbol{c} . \end{aligned}R(A+a,B)C=R(A,B)C+R(a,B)CR(A,B+b)C=R(A,B)C+R(A,b)CR,A,B)(C+c)=R(A,B)C+R(A,B)c.
(d) Now use the above properties to prove the most crucial feature of R ( A , B ) C R ( A , B ) C R(A,B)C\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}R(A,B)C : Modify the variations (gradients) of A , B A , B A,B\boldsymbol{A}, \boldsymbol{B}A,B, and C C C\boldsymbol{C}C in an arbitrary manner, but leave A , B , C A , B , C A,B,C\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}A,B,C unchanged at P 0 P 0 P_(0)\mathscr{P}_{0}P0 :
A A + a α e α B B + b α e α C C + c α e α } a α ( P ) , b α ( P ) , c α ( P ) arbitrary except they all vanish at P = P 0 . A A + a α e α B B + b α e α C C + c α e α a α ( P ) , b α ( P ) , c α ( P )  arbitrary except   they all vanish at  P = P 0 . {:[A longrightarrow A+a^(alpha)e_(alpha)],[B longrightarrow B+b^(alpha)e_(alpha)],[C longrightarrow C+c^(alpha)e_(alpha)]}quad{:[a^(alpha)(P)","b^(alpha)(P)","c^(alpha)(P)" arbitrary except "],[" they all vanish at "P=P_(0).]:}\left.\begin{array}{l} \boldsymbol{A} \longrightarrow \boldsymbol{A}+a^{\alpha} \boldsymbol{e}_{\alpha} \\ \boldsymbol{B} \longrightarrow \boldsymbol{B}+b^{\alpha} \boldsymbol{e}_{\alpha} \\ \boldsymbol{C} \longrightarrow \boldsymbol{C}+c^{\alpha} \boldsymbol{e}_{\alpha} \end{array}\right\} \quad \begin{aligned} & a^{\alpha}(\mathscr{P}), b^{\alpha}(\mathscr{P}), c^{\alpha}(\mathscr{P}) \text { arbitrary except } \\ & \text { they all vanish at } \mathscr{P}=\mathscr{P}_{0} . \end{aligned}AA+aαeαBB+bαeαCC+cαeα}aα(P),bα(P),cα(P) arbitrary except  they all vanish at P=P0.
Show that this modification leaves R ( A , B ) C R ( A , B ) C R(A,B)C\mathscr{R}(\boldsymbol{A}, \boldsymbol{B}) \boldsymbol{C}R(A,B)C unchanged at P 0 P 0 P_(0)\mathscr{P}_{0}P0. (e) From these facts, conclude that Riemann is a tensor.

Exercise 11.3. COMPONENTS OF Riemann IN COORDINATE BASIS

Derive equation (11.12) for the components of the Riemann tensor in a coordinate basis. [Solution:
R α β γ δ = R i e m a n n ( ω α , e β , e γ , e δ ) [ standard way to calculate components ] = ω α , R ( e γ , e δ ) e β [by definition (11.9)] = ω α , ( γ δ δ γ ) e β [ by definition (11.8) plus [ e γ , e δ ] = 0 in coord. basis ] = ω α , e μ Γ μ β δ , γ + ( e ν Γ ν μ γ ) Γ μ β δ e μ Γ μ β γ , δ ( e ν Γ v μ δ ) Γ μ β γ = ( Γ μ β δ , γ Γ μ β γ , δ , ) ω α , e μ + ( Γ ν μ γ Γ μ β δ Γ ν μ δ Γ μ β γ ) ω α , e γ , R α β γ δ = R i e m a n n ω α , e β , e γ , e δ  standard way to   calculate components  = ω α , R e γ , e δ e β  [by definition (11.9)]  = ω α , γ δ δ γ e β  by definition (11.8) plus  e γ , e δ = 0  in coord. basis  = ω α , e μ Γ μ β δ , γ + e ν Γ ν μ γ Γ μ β δ e μ Γ μ β γ , δ e ν Γ v μ δ Γ μ β γ = Γ μ β δ , γ Γ μ β γ , δ , ω α , e μ + Γ ν μ γ Γ μ β δ Γ ν μ δ Γ μ β γ ω α , e γ , {:[R^(alpha)_(beta gamma delta)=Riemann(omega^(alpha),e_(beta),e_(gamma),e_(delta))quad[[" standard way to "],[" calculate components "]]],[=(:omega^(alpha),R(e_(gamma),e_(delta))e_(beta):)quad" [by definition (11.9)] "],[=(:omega^(alpha),(grad_(gamma)grad_(delta)-grad_(delta)grad_(gamma))e_(beta):)quad[[" by definition (11.8) plus "],[[e_(gamma),e_(delta)]=0" in coord. basis "]]],[=(:omega^(alpha),e_(mu)Gamma^(mu)_(beta delta,gamma)+(e_(nu)Gamma^(nu)_(mu gamma))Gamma^(mu)_(beta delta)-e_(mu)Gamma^(mu)_(beta gamma,delta)-(e_(nu)Gamma^(v)_(mu delta))Gamma^(mu)_(beta gamma):)],[=(Gamma^(mu)_(beta delta,gamma)-Gamma^(mu)_(beta gamma,delta,))(:omega^(alpha),e_(mu):)+(Gamma^(nu)_(mu gamma)Gamma^(mu)_(beta delta)-Gamma^(nu)_(mu delta)Gamma^(mu)_(beta gamma))(:omega^(alpha),e_(gamma):)","]:}\begin{aligned} & R^{\alpha}{ }_{\beta \gamma \delta}=\boldsymbol{R i e m a n n}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \quad\left[\begin{array}{l} \text { standard way to } \\ \text { calculate components } \end{array}\right] \\ & =\left\langle\boldsymbol{\omega}^{\alpha}, \mathscr{R}\left(\boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \boldsymbol{e}_{\beta}\right\rangle \quad \text { [by definition (11.9)] } \\ & =\left\langle\boldsymbol{\omega}^{\alpha},\left(\boldsymbol{\nabla}_{\gamma} \boldsymbol{\nabla}_{\delta}-\boldsymbol{\nabla}_{\delta} \boldsymbol{\nabla}_{\gamma}\right) \boldsymbol{e}_{\beta}\right\rangle \quad\left[\begin{array}{l} \text { by definition (11.8) plus } \\ {\left[\boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right]=0 \text { in coord. basis }} \end{array}\right] \\ & =\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\mu} \Gamma^{\mu}{ }_{\beta \delta, \gamma}+\left(\boldsymbol{e}_{\nu} \Gamma^{\nu}{ }_{\mu \gamma}\right) \Gamma^{\mu}{ }_{\beta \delta}-\boldsymbol{e}_{\mu} \Gamma^{\mu}{ }_{\beta \gamma, \delta}-\left(\boldsymbol{e}_{\nu} \Gamma^{v}{ }_{\mu \delta}\right) \Gamma^{\mu}{ }_{\beta \gamma}\right\rangle \\ & =\left(\Gamma^{\mu}{ }_{\beta \delta, \gamma}-\Gamma^{\mu}{ }_{\beta \gamma, \delta,}\right)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\mu}\right\rangle+\left(\Gamma^{\nu}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta}-\Gamma^{\nu}{ }_{\mu \delta} \Gamma^{\mu}{ }_{\beta \gamma}\right)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\gamma}\right\rangle, \end{aligned}Rαβγδ=Riemann(ωα,eβ,eγ,eδ)[ standard way to  calculate components ]=ωα,R(eγ,eδ)eβ [by definition (11.9)] =ωα,(γδδγ)eβ[ by definition (11.8) plus [eγ,eδ]=0 in coord. basis ]=ωα,eμΓμβδ,γ+(eνΓνμγ)ΓμβδeμΓμβγ,δ(eνΓvμδ)Γμβγ=(Γμβδ,γΓμβγ,δ,)ωα,eμ+(ΓνμγΓμβδΓνμδΓμβγ)ωα,eγ,
which reduces (upon using ω α , e μ = δ α μ ω α , e μ = δ α μ (:omega^(alpha),e_(mu):)=delta^(alpha)_(mu)\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\mu}\right\rangle=\delta^{\alpha}{ }_{\mu}ωα,eμ=δαμ ) to (11.12).]

Exercise 11.4. COMPONENTS OF RIEMANN IN NONCOORDINATE BASIS

In a noncoordinate basis with commutation coefficients c α β γ c α β γ c_(alpha beta)^(gamma){c_{\alpha \beta}}^{\gamma}cαβγ defined by equation (9.22), derive the following equation for the components of Riemann:
(11.13) R α β γ δ = Γ α β 8 , γ Γ α β γ , δ + Γ α μ γ Γ μ β δ Γ α μ δ Γ μ β γ Γ α β μ c γ δ μ . (11.13) R α β γ δ = Γ α β 8 , γ Γ α β γ , δ + Γ α μ γ Γ μ β δ Γ α μ δ Γ μ β γ Γ α β μ c γ δ μ . {:(11.13)R^(alpha)_(beta gamma delta)=Gamma^(alpha)_(beta8,gamma)-Gamma^(alpha)_(beta gamma,delta)+Gamma^(alpha)_(mu gamma)Gamma^(mu)_(beta delta)-Gamma^(alpha)_(mu delta)Gamma^(mu)_(beta gamma)-Gamma^(alpha)_(beta mu)c_(gamma delta)^(mu).:}\begin{equation*} R^{\alpha}{ }_{\beta \gamma \delta}=\Gamma^{\alpha}{ }_{\beta 8, \gamma}-\Gamma^{\alpha}{ }_{\beta \gamma, \delta}+\Gamma^{\alpha}{ }_{\mu \gamma} \Gamma^{\mu}{ }_{\beta \delta}-\Gamma^{\alpha}{ }_{\mu \delta} \Gamma^{\mu}{ }_{\beta \gamma}-\Gamma^{\alpha}{ }_{\beta \mu} c_{\gamma \delta}{ }^{\mu} . \tag{11.13} \end{equation*}(11.13)Rαβγδ=Γαβ8,γΓαβγ,δ+ΓαμγΓμβδΓαμδΓμβγΓαβμcγδμ.

§11.4. PARALLEL TRANSPORT AROUND A CLOSED CURVE

What are the effects of spacetime curvature, and how can one quantify them? One effect is geodesic deviation (relative acceleration of test bodies), quantified by equation (11.10). Another effect, almost as important, is the change in a vector caused by parallel transport around a closed curve. This effect shows up most clearly in the same problem, geodesic deviation, that motivated curvature in the first place. The relative acceleration vector u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}uun is also the change δ u δ u delta u\delta \boldsymbol{u}δu in the vector u u u\boldsymbol{u}u caused by parallel transport around the curve whose legs are the vectors n n n\boldsymbol{n}n and u u u\boldsymbol{u}u :
u u n = δ u u u n = δ u grad_(u)grad_(u)n=delta u\nabla_{u} \nabla_{u} n=\delta uuun=δu
(See Box 11.6 for proof.) Hence, in this special case one can write
δ u + Riemann ( , u , n , u ) = 0 δ u + Riemann ( , u , n , u ) = 0 delta u+Riemann(dots,u,n,u)=0\delta u+\operatorname{Riemann}(\ldots, u, n, u)=0δu+Riemann(,u,n,u)=0
The expected generalization is obvious: pick a closed quadrilateral with legs u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa and v Δ b v Δ b v Delta b\boldsymbol{v} \Delta bvΔb (Figure 11.2; Δ a Δ a Delta a\Delta aΔa and Δ b Δ b Delta b\Delta bΔb are small parameters, to go to zero at end of discussion). Parallel transport the vector A A A\boldsymbol{A}A around this quadrilateral. The resultant change in A A A\boldsymbol{A}A should satisfy the equation
(11.14) δ A + R i e m a n n ( , A , u Δ a , v Δ b ) = 0 (11.14) δ A + R i e m a n n ( , A , u Δ a , v Δ b ) = 0 {:(11.14)delta A+Riemann(dots","A","u Delta a","v Delta b)=0:}\begin{equation*} \delta \boldsymbol{A}+\boldsymbol{R i e m a n n}(\ldots, \boldsymbol{A}, \boldsymbol{u} \Delta a, \boldsymbol{v} \Delta b)=0 \tag{11.14} \end{equation*}(11.14)δA+Riemann(,A,uΔa,vΔb)=0
or, equivalently,
(11.14') δ A + Δ a Δ b ( u , v ) A = 0 (11.14') δ A + Δ a Δ b ( u , v ) A = 0 {:(11.14')delta A+Delta a Delta bℜ(u","v)A=0:}\begin{equation*} \delta \boldsymbol{A}+\Delta a \Delta b \Re(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{A}=0 \tag{11.14'} \end{equation*}(11.14')δA+ΔaΔb(u,v)A=0
or, more precisely,
(11.14") Lim Δ a 0 Δ b 0 ( δ A Δ a Δ b ) + Riemann ( , A , u , v ) = 0 (11.14") Lim Δ a 0 Δ b 0 δ A Δ a Δ b +  Riemann  ( , A , u , v ) = 0 {:(11.14)Lim_({:[Delta a rarr0],[Delta b rarr0]:})((delta A)/(Delta a Delta b))+" Riemann "(dots","A","u","v)=0:}\begin{equation*} \operatorname{Lim}_{\substack{\Delta a \rightarrow 0 \\ \Delta b \rightarrow 0}}\left(\frac{\delta \boldsymbol{A}}{\Delta a \Delta b}\right)+\text { Riemann }(\ldots, \boldsymbol{A}, \boldsymbol{u}, \boldsymbol{v})=0 \tag{11.14"} \end{equation*}(11.14")LimΔa0Δb0(δAΔaΔb)+ Riemann (,A,u,v)=0
The proof is enlightening, for it reveals the geometric origin of the correction term [ u , v ] [ u , v ] grad_([u,v])\nabla_{[u, v]}[u,v] in the curvature operator.
The circuit of transport (Figure 11.2) is to be made from two arbitrary vector fields u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa and v Δ b v Δ b v Delta b\boldsymbol{v} \Delta bvΔb. However, a circuit made only of these fields has a gap in it, for a simple reason. The magnitude of u u u\boldsymbol{u}u varies the wrong way from place to place. The displacement u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa that reaches across at the bottom of the quadrilateral from
Change in a vector due to parallel transport around a closed curve:
Related to geodesic deviation
Equation for change
"
Figure 11.2.
The change δ A δ A delta A\delta \boldsymbol{A}δA in a vector A A A\boldsymbol{A}A as a result of parallel transport around a closed curve. The edges of the curve are the vector fields u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa and v Δ b v Δ b v Delta b\boldsymbol{v} \Delta bvΔb, plus the "closer of the quadrilateral" [ v Δ b , u Δ a ] = [ v , u ] Δ a [ v Δ b , u Δ a ] = [ v , u ] Δ a [v Delta b,u Delta a]=[v,u]Delta a[\boldsymbol{v} \Delta b, \boldsymbol{u} \Delta a]=[\boldsymbol{v}, \boldsymbol{u}] \Delta a[vΔb,uΔa]=[v,u]Δa Δ b Δ b Delta b\Delta bΔb (see Box 9.2).
one line of v v v\boldsymbol{v}v 's to another cannot make the connection at the top of the quadrilateral. Similarly the v v v\boldsymbol{v}v 's vary the wrong way from place to place to connect the u u u\boldsymbol{u}u 's. To close the gap and complete the circuit, insert the "closer of quadrilaterals" [ v Δ b [ v Δ b [vDelta b[\mathbf{v} \Delta b[vΔb, u Δ a ] = [ v , u ] Δ a Δ b u Δ a ] = [ v , u ] Δ a Δ b u Delta a]=[v,u]Delta a Delta b\boldsymbol{u} \Delta a]=[\boldsymbol{v}, \boldsymbol{u}] \Delta a \Delta buΔa]=[v,u]ΔaΔb. (See Box 9.2 for why this vector closes the gap.)
With the route now specified, the vector A A A\boldsymbol{A}A is to be transported around it. One way to do this, "geometrical construction" by the method of Schild's ladder applied over and over, is the foundation for planning a possible experiment. For planning an abstract and coordinate-free calculation (the present line of action), introduce a "fiducial field," only to take it away at the end of the calculation. Plan: Conceive of A A A\boldsymbol{A}A, not as a localized vector defined solely at the start of the trip, but as a vector field (defined throughout the trip). Purpose: To provide a standard of reference (comparison of A A A\boldsymbol{A}A transported from the origin with A A A\boldsymbol{A}A at the place in question). Principle: The standard of reference will cancel out in the end. Procedure:
δ A = ( Net change made in taking the vector A , originally localized at the start of the circuit, and transporting it parallel to itself ("mobile A ") around the closed circuit. This quantity cannot be evaluated until completion of circuit because there is no preexisting standard of reference along the way. ) δ A =  Net change made in taking the vector  A , originally localized at the   start of the circuit, and transporting it parallel to itself ("mobile  A  ")   around the closed circuit. This quantity cannot be evaluated until   completion of circuit because there is no preexisting standard of   reference along the way.  -delta A=-([" Net change made in taking the vector "A", originally localized at the "],[" start of the circuit, and transporting it parallel to itself ("mobile "A" ") "],[" around the closed circuit. This quantity cannot be evaluated until "],[" completion of circuit because there is no preexisting standard of "],[" reference along the way. "])-\delta \boldsymbol{A}=-\left(\begin{array}{l}\text { Net change made in taking the vector } \boldsymbol{A} \text {, originally localized at the } \\ \text { start of the circuit, and transporting it parallel to itself ("mobile } \boldsymbol{A} \text { ") } \\ \text { around the closed circuit. This quantity cannot be evaluated until } \\ \text { completion of circuit because there is no preexisting standard of } \\ \text { reference along the way. }\end{array}\right)δA=( Net change made in taking the vector A, originally localized at the  start of the circuit, and transporting it parallel to itself ("mobile A ")  around the closed circuit. This quantity cannot be evaluated until  completion of circuit because there is no preexisting standard of  reference along the way. )
= + ( A quantity subject to analysis for each leg of circuit individually. This new quantity is defined by introducing throughout the whole region a vector field A (field) , smoothly varying, and in agreement at starting point with the original localized A , but otherwise arbitrary. This new quantity is then given by A (field) at starting point (same as A (localized) at starting point) minus A (mobile) ) at finish point (after transit). ) = +  A quantity subject to analysis for each leg of circuit individually. This   new quantity is defined by introducing throughout the whole region   a vector field  A (field)  , smoothly varying, and in agreement at starting   point with the original localized  A ,  but otherwise arbitrary. This new   quantity is then given by  A (field)   at starting point (same as  A (localized)   at starting point) minus  A (mobile)  )  at finish point (after transit).  =+([" A quantity subject to analysis for each leg of circuit individually. This "],[" new quantity is defined by introducing throughout the whole region "],[" a vector field "A^((field) )", smoothly varying, and in agreement at starting "],[" point with the original localized "A","" but otherwise arbitrary. This new "],[" quantity is then given by "A^((field) )" at starting point (same as "A^((localized) )],[" at starting point) minus "A^((mobile) ))" at finish point (after transit). "])=+\left(\begin{array}{l}\text { A quantity subject to analysis for each leg of circuit individually. This } \\ \text { new quantity is defined by introducing throughout the whole region } \\ \text { a vector field } \boldsymbol{A}^{\text {(field) }} \text {, smoothly varying, and in agreement at starting } \\ \text { point with the original localized } \boldsymbol{A}, \text { but otherwise arbitrary. This new } \\ \text { quantity is then given by } \boldsymbol{A}^{\text {(field) }} \text { at starting point (same as } \boldsymbol{A}^{\text {(localized) }} \\ \text { at starting point) minus } \boldsymbol{A}^{\text {(mobile) })} \text { at finish point (after transit). }\end{array}\right)=+( A quantity subject to analysis for each leg of circuit individually. This  new quantity is defined by introducing throughout the whole region  a vector field A(field) , smoothly varying, and in agreement at starting  point with the original localized A, but otherwise arbitrary. This new  quantity is then given by A(field)  at starting point (same as A(localized)  at starting point) minus A(mobile) ) at finish point (after transit). )
= legs of circuit ( Change in A ( field ) relative to A ( mobile) in the course of transport along specified leg. Value for any one leg depends on the arbitrary choice of A ( field ) , but this arbitrariness cancels out in end because of closure of circuit. ) =  legs of   circuit   Change in  A ( field  )  relative to  A ( mobile)   in the course of transport along   specified leg. Value for any one leg depends on the arbitrary choice   of  A ( field  ) , but this arbitrariness cancels out in end because of closure   of circuit.  =sum_({:[" legs of "],[" circuit "]:})([" Change in "A^(("field "))" relative to "A^(("mobile) ")" in the course of transport along "],[" specified leg. Value for any one leg depends on the arbitrary choice "],[" of "A^(("field "))", but this arbitrariness cancels out in end because of closure "],[" of circuit. "])=\sum_{\substack{\text { legs of } \\ \text { circuit }}}\left(\begin{array}{l}\text { Change in } \boldsymbol{A}^{(\text {field })} \text { relative to } \boldsymbol{A}^{(\text {mobile) }} \text { in the course of transport along } \\ \text { specified leg. Value for any one leg depends on the arbitrary choice } \\ \text { of } \boldsymbol{A}^{(\text {field })} \text {, but this arbitrariness cancels out in end because of closure } \\ \text { of circuit. }\end{array}\right)= legs of  circuit ( Change in A(field ) relative to A(mobile)  in the course of transport along  specified leg. Value for any one leg depends on the arbitrary choice  of A(field ), but this arbitrariness cancels out in end because of closure  of circuit. )
= ( Change in A (field) relative to the parallel-transported A (mobile) as standard of reference, made up of contributions along following legs of Figure 11.2 : v Δ b , giving v A (field) Δ b (on line displaced u Δ a from start) v Δ b , giving v A (field) Δ b (on line through starting point) u Δ a , giving u A (field) Δ a (on line displaced v Δ b from start) + u Δ a , giving + [ u u A (field) Δ a (on line through starting point) + v , u ] Δ a Δ b , giving [v,.u] A (field) Δ a Δ b ) = { u v v u + [ v , u ] } A ( field ) Δ a Δ b (11.15) = R i e m a n n ( , A (field ) , u , v ) Δ a Δ b = R ( u , v ) A (field) Δ a Δ b . =  Change in  A (field)   relative to the parallel-transported  A (mobile)   as standard   of reference, made up of contributions along following legs of Figure  11.2  :  v Δ b , giving  v A (field)  Δ b  (on line displaced  u Δ a  from start)  v Δ b , giving  v A (field)  Δ b  (on line through starting point)  u Δ a , giving  u A (field)  Δ a  (on line displaced  v Δ b  from start)  + u Δ a , giving  + u u A (field)  Δ a  (on line through starting point)  + v , u ] Δ a Δ b ,  giving  [v,.u]  A (field)  Δ a Δ b = u v v u + [ v , u ] A ( field  ) Δ a Δ b (11.15) = R i e m a n n , A (field  ) , u , v Δ a Δ b = R ( u , v ) A (field)  Δ a Δ b {:[=([" Change in "A^((field) )" relative to the parallel-transported "A^((mobile) )" as standard "],[" of reference, made up of contributions along following legs of Figure "],[11.2" : "],[v Delta b", giving "quadgrad_(v)A^((field) )Delta b" (on line displaced "u Delta a" from start) "],[-v Delta b", giving "-grad_(v)A^((field) )Delta b" (on line through starting point) "],[-u Delta a", giving "-grad_(u)A^((field) )Delta a" (on line displaced "v Delta b" from start) "],[+u Delta a", giving "],[+[u_(u)A^((field) )Delta a:}" (on line through starting point) "],[+v","u]Delta a Delta b","" giving "grad_("[v,.u] ")A^((field) )Delta a Delta b])],[={grad_(u)grad_(v)-grad_(v)grad_(u)+grad_([v,u])}A^(("field "))Delta a Delta b],[(11.15)=Riemann(dots,A^((field )),u,v)Delta a Delta b=R(u","v)A^((field) )Delta a Delta b". "]:}\begin{align*} & =\left(\begin{array}{l} \text { Change in } \boldsymbol{A}^{\text {(field) }} \text { relative to the parallel-transported } \boldsymbol{A}^{\text {(mobile) }} \text { as standard } \\ \text { of reference, made up of contributions along following legs of Figure } \\ 11.2 \text { : } \\ \boldsymbol{v} \Delta b \text {, giving } \quad \boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{A}^{\text {(field) }} \Delta b \text { (on line displaced } \boldsymbol{u} \Delta a \text { from start) } \\ -\boldsymbol{v} \Delta b \text {, giving }-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{A}^{\text {(field) }} \Delta b \text { (on line through starting point) } \\ -\boldsymbol{u} \Delta a \text {, giving }-\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{A}^{\text {(field) }} \Delta a \text { (on line displaced } \boldsymbol{v} \Delta b \text { from start) } \\ +\boldsymbol{u} \Delta a \text {, giving } \\ +\left[\boldsymbol{\boldsymbol { u } _ { \boldsymbol { u } }} \boldsymbol{A}^{\text {(field) }} \Delta a\right. \text { (on line through starting point) } \\ +\boldsymbol{v}, \boldsymbol{u}] \Delta a \Delta b, \text { giving } \boldsymbol{\nabla}_{\text {[v,.u] }} \boldsymbol{A}^{\text {(field) }} \Delta a \Delta b \end{array}\right) \\ & =\left\{\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{v}-\boldsymbol{\nabla}_{v} \boldsymbol{\nabla}_{u}+\boldsymbol{\nabla}_{[v, u]}\right\} \boldsymbol{A}^{(\text {field })} \Delta a \Delta b \\ & =\boldsymbol{R i e m a n n}\left(\ldots, \boldsymbol{A}^{\text {(field })}, \boldsymbol{u}, \boldsymbol{v}\right) \Delta a \Delta b=\mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{A}^{\text {(field) }} \Delta a \Delta b \text {. } \tag{11.15} \end{align*}=( Change in A(field)  relative to the parallel-transported A(mobile)  as standard  of reference, made up of contributions along following legs of Figure 11.2 : vΔb, giving vA(field) Δb (on line displaced uΔa from start) vΔb, giving vA(field) Δb (on line through starting point) uΔa, giving uA(field) Δa (on line displaced vΔb from start) +uΔa, giving +[uuA(field) Δa (on line through starting point) +v,u]ΔaΔb, giving [v,.u] A(field) ΔaΔb)={uvvu+[v,u]}A(field )ΔaΔb(11.15)=Riemann(,A(field ),u,v)ΔaΔb=R(u,v)A(field) ΔaΔb
Profit: The curvature operator
Riemann ( , , u , v ) = R ( u , v ) = [ u , v ] [ u , v ] Riemann ( , , u , v ) = R ( u , v ) = u , v [ u , v ] Riemann(dots,dots,u,v)=R(u,v)=[grad_(u),grad_(v)]-grad_([u,v])\operatorname{Riemann}(\ldots, \ldots, \boldsymbol{u}, \boldsymbol{v})=\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})=\left[\boldsymbol{\nabla}_{u}, \boldsymbol{\nabla}_{v}\right]-\boldsymbol{\nabla}_{[u, v]}Riemann(,,u,v)=R(u,v)=[u,v][u,v]

Box 11.6 GEODESIC DEVIATION AND PARALLEL TRANSPORT AROUND closed curve: two aspects of same construction

Geodesic Deviation

u u n = Lim Δ λ 0 Δ n 0 { a P + B R ( Δ λ ) 2 Δ n } . u u n = Lim Δ λ 0 Δ n 0 a P + B R ( Δ λ ) 2 Δ n . grad_(u)grad_(u)n=Lim_({:[Delta lambda rarr0],[Delta n rarr0]:}){(aP+BR)/((Delta lambda)^(2)Delta n)}.\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} \boldsymbol{n}=\operatorname{Lim}_{\substack{\Delta \lambda \rightarrow 0 \\ \Delta n \rightarrow 0}}\left\{\frac{a \mathscr{P}+\mathscr{B} \mathscr{R}}{(\Delta \lambda)^{2} \Delta n}\right\} .uun=LimΔλ0Δn0{aP+BR(Δλ)2Δn}.
(See Boxes 11.2 and 11.3)

Geodesic Deviation

Same result; different construction. To simplify the connection with closed-curve transport, change the tilt and dilate the parametrization of geodesic P R P R PR\mathscr{P} \mathscr{R}PR in A A A\mathbf{A}A. The result: B, where P P ¯ bar(P)\overline{\mathscr{P}}P and a a aaa coincide. From F F F\mathbf{F}F of Box 11.2 one knows C P + C P + CP+\mathbb{C P}+CP+ B B = a P + B R B B = a P ¯ + B R ¯ BB=a bar(P)+B bar(R)\mathscr{B} \mathscr{B}=\mathscr{a} \overline{\mathscr{P}}+\mathscr{B} \overline{\mathscr{R}}BB=aP+BR-i.e. u u n u u n quadgrad_(u)grad_(u)n\quad \boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} \boldsymbol{n}uun is the
Box 11.6 (continued)
same for this family of geodesics as for the original family
u u n = Lim Δ λ 0 Δ n 0 { B R ¯ ( Δ λ ) 2 Δ n } . u u n = Lim Δ λ 0 Δ n 0 B R ¯ ( Δ λ ) 2 Δ n . grad_(u)grad_(u)n=Lim_({:[Delta lambda rarr0],[Delta n rarr0]:}){(B( bar(R)))/((Delta lambda)^(2)Delta n)}.\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}=\operatorname{Lim}_{\substack{\Delta \lambda \rightarrow 0 \\ \Delta n \rightarrow 0}}\left\{\frac{\mathscr{B} \bar{R}}{(\Delta \lambda)^{2} \Delta n}\right\} .uun=LimΔλ0Δn0{BR¯(Δλ)2Δn}.
Also, to simplify discussion set Δ n = Δ λ = 1 Δ n = Δ λ = 1 Delta n=Delta lambda=1\Delta n=\Delta \lambda=1Δn=Δλ=1, and assume n n n\boldsymbol{n}n and u u u\boldsymbol{u}u are small enough that one can evaluate u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}uun without taking the limit:
u u n = B ¯ . u u n = B ¯ . grad_(u)grad_(u)n=B bar(ℜ).\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} n=\mathscr{B} \bar{\Re} .uun=B¯.

Parallel Transport Around Closed Curve, Performed by Same Construction

Plan: Parallel transport the vector u Δ λ = Q ¯ u Δ λ = Q ¯ u Delta lambda=Q bar(ℜ)\boldsymbol{u} \Delta \lambda=\mathcal{Q} \bar{\Re}uΔλ=Q¯ counterclockwise around the curve Q P L K Q P ¯ L K Qlongrightarrow bar(P)longrightarrowLlongrightarrowKlongrightarrow\mathscr{Q} \longrightarrow \overline{\mathscr{P}} \longrightarrow \mathcal{L} \longrightarrow \mathscr{K} \longrightarrowQPLK 2. Execution: (1) Call transported vector u ( m ) u ( m ) u^((m))\boldsymbol{u}^{(\mathrm{m})}u(m) ("m" for "mobile"). (2) At Q , u ( m ) = Q ¯ Q , u ( m ) = Q ¯ Q,u^((m))=Q bar(ℜ)\mathcal{Q}, \boldsymbol{u}^{(\mathrm{m})}=\mathcal{Q} \bar{\Re}Q,u(m)=Q¯. (3) At P , u ( m ) = P Q P ¯ , u ( m ) = P ¯ Q bar(P),u^((m))= bar(P)Q\overline{\mathscr{P}}, \boldsymbol{u}^{(\mathrm{m})}=\overline{\mathscr{P}} \mathcal{Q}P,u(m)=PQ because P Q R P ¯ Q R ¯ bar(P)Q bar(R)\overline{\mathcal{P}} \mathcal{Q} \overline{\mathcal{R}}PQR is a geodesic and u ( m ) u ( m ) u^((m))\boldsymbol{u}^{(\mathrm{m})}u(m) is its tangent vector. (4) At L , u ( m ) = E R L , u ( m ) = E R L,u^((m))=ER\mathcal{L}, \boldsymbol{u}^{(\mathrm{m})}=\mathfrak{E} \mathscr{R}L,u(m)=ER according to Schild's ladder of the picture. (5) At , u ( m ) = R , u ( m ) = R ℜ,u^((m))=R\Re, \boldsymbol{u}^{(\mathrm{m})}=\mathscr{R},u(m)=R because R R R R RR\mathscr{R} \mathscr{R}RR is a geodesic and u ( m ) u ( m ) u^((m))\boldsymbol{u}^{(\mathrm{m})}u(m) is now its tangent vector. (6) At Q , u ( m ) = 2 2 Q , u ( m ) = 2 2 Q,u^((m))=22\mathcal{Q}, \boldsymbol{u}^{(\mathrm{m})}=2 \mathscr{2}Q,u(m)=22 according to Schild's ladder. Result: The change in u ( m ) u ( m ) u^((m))\boldsymbol{u}^{(\mathrm{m})}u(m) is B R B R ¯ -B bar(R)-\mathscr{B} \overline{\mathfrak{R}}BR. Had the curve been circuited in opposite direction ( E P ( E P ¯ (Elongrightarrow bar(P)longrightarrow(\mathcal{E} \longrightarrow \overline{\mathscr{P}} \longrightarrow(EP Q R R Q R R QlongrightarrowRlongrightarrowR\mathscr{Q} \longrightarrow \mathscr{R} \longrightarrow \mathcal{R}QRR ), the change would have been + B R : + B R ¯ : +B bar(R):+\mathscr{B} \overline{\mathscr{R}}:+BR:
( δ u ) due to parallel transport up n , out u , down n , and = B R = u u n . back along u to starting point ( δ u ) due to parallel transport up  n ,  out  u ,  down  n ,  and  = B R ¯ = u u n .  back along  u  to starting point  {:[(delta u)_("due to parallel transport up "n," out "u," down "-n," and ")=B bar(R)=grad_(u)grad_(u)n.],[quad" back along "-u" to starting point "]:}\begin{aligned} & (\delta \boldsymbol{u})_{\text {due to parallel transport up } n, \text { out } \boldsymbol{u}, \text { down }-\boldsymbol{n}, \text { and }}=\mathscr{B} \overline{\mathscr{R}}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n} . \\ & \quad \text { back along }-\boldsymbol{u} \text { to starting point } \end{aligned}(δu)due to parallel transport up n, out u, down n, and =BR=uun. back along u to starting point 
applied to the vector field A (field) A (field)  A^((field) )\boldsymbol{A}^{\text {(field) }}A(field) , gives the negative of the change in the localized vector A (localized) A (localized)  A^((localized) )\boldsymbol{A}^{\text {(localized) }}A(localized)  (called A (mobile) A (mobile)  A^((mobile) )\boldsymbol{A}^{\text {(mobile) }}A(mobile)  during the phase of travel) on parallel transport around the closed circuit. It does not give the change in A (field ) A (field  ) A^((field ))\boldsymbol{A}^{\text {(field })}A(field ) on traversal of that circuit, for A (field) A (field)  A^((field) )\boldsymbol{A}^{\text {(field) }}A(field)  has the same value at the end of the journey as at the beginning. Equation (11.14') expresses that change in terms of the conveniently calculated differential operator, R ( u , v ) = [ u , v ] [ u , v ] R ( u , v ) = u , v [ u , v ] R(u,v)=[grad_(u),grad_(v)]-grad_([u,v])\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})=\left[\boldsymbol{\nabla}_{\boldsymbol{u}}, \boldsymbol{\nabla}_{\mathbf{v}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{u}, \boldsymbol{v}]}R(u,v)=[u,v][u,v]. Paradox: Neither wanted nor evaluated is the change in the quantity A (field) A (field)  A^((field) )\boldsymbol{A}^{\text {(field) }}A(field)  acted on by this operator. Payoff: Ostensibly differential in the character of its action on A A A\boldsymbol{A}A, the operator Riemann ( , , u , v ) = R ( u , v ) ( , , u , v ) = R ( u , v ) (dots,dots,u,v)=R(u,v)(\ldots, \ldots, \boldsymbol{u}, \boldsymbol{v})=\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})(,,u,v)=R(u,v) is actually local. Thus, replace the proposed smoothly varying vector field A ( field ) A ( field  ) A^(("field "))\boldsymbol{A}^{(\text {field })}A(field ) by a quite different but also smoothly varying vector field A (rield, new) A (rield, new)  A^((rield, new) )\boldsymbol{A}^{\text {(rield, new) }}A(rield, new) . Then the two fields need agree only at the one point in question for them to give the same output Riemann ( , A , u , v ) = R ( u , v ) A ( , A , u , v ) = R ( u , v ) A (dots,A,u,v)=R(u,v)A(\ldots, \boldsymbol{A}, \boldsymbol{u}, \boldsymbol{v})=\mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{A}(,A,u,v)=R(u,v)A at that point. This one
knows from the fact that δ A δ A delta A\delta \boldsymbol{A}δA, the quantity calculated, has an existence and value independent of the choice of A (field). A (field).  A^((field). )\boldsymbol{A}^{\text {(field). }}A(field). . This one can also verify by detailed calculation (exercise 11.2). Power: Although they cancel out in their response to any change of A A A\boldsymbol{A}A with location, the several differentiations in the curvature operator respond directly to the "rate of change of geometry with location" ("geodesic deviation"). Prolongation: The closed curve need not be a quadrilateral. The curvature operator tells how a vector changes on parallel transport about small curves of arbitrary shape (Box 11.7).

Exercise 11.5. COPLANARITY OF CLOSED CURVES

EXERCISE

Let f 1 f 1 f_(1)\boldsymbol{f}_{1}f1 and f 2 f 2 f_(2)\boldsymbol{f}_{2}f2 be the bivectors (see Box 11.7) for two small closed curves at the same event. Show that the curves are coplanar if and only if f 1 = a f 2 f 1 = a f 2 f_(1)=af_(2)\boldsymbol{f}_{1}=a \boldsymbol{f}_{2}f1=af2 for some number a a aaa.
Box 11.7 THE LAW FOR PARALLEL TRANSPORT ABOUT A CLOSED CURVE

A. Special Case

Curve is closed quadrilateral formed by vector fields u u u\boldsymbol{u}u and v v v\boldsymbol{v}v.
  1. Law says (in component form)
(1) δ A α + R α β γ δ A β u γ v δ = 0 . (1) δ A α + R α β γ δ A β u γ v δ = 0 . {:(1)deltaA^(alpha)+R^(alpha)_(beta gamma delta)A^(beta)u^(gamma)v^(delta)=0.:}\begin{equation*} \delta A^{\alpha}+R^{\alpha}{ }_{\beta \gamma \delta} A^{\beta} u^{\gamma} v^{\delta}=0 . \tag{1} \end{equation*}(1)δAα+RαβγδAβuγvδ=0.
  1. On what characteristics of the closed curve does this depend?
    a. Notice that R α β γ δ = R α β δ γ R α β γ δ = R α β δ γ R^(alpha)_(beta gamma delta)=-R^(alpha)_(beta delta gamma)R^{\alpha}{ }_{\beta \gamma \delta}=-R^{\alpha}{ }_{\beta \delta \gamma}Rαβγδ=Rαβδγ (antisymmetry in last two indices; obvious in equation 11.12 for components; also obvious because reversing the direction the curve is traversed-i.e., interchanging u u u\boldsymbol{u}u and v v v\boldsymbol{v}v-should reverse sign of δ A δ A delta A\delta \boldsymbol{A}δA ).
    b. Equation (1) contracts u v u v u ox v\boldsymbol{u} \otimes \boldsymbol{v}uv into these antisymmetric, last two indices. The symmetric part of u v u v u ox v\boldsymbol{u} \otimes \boldsymbol{v}uv must give zero. Only the antisymmetric part, u v = u v v u u v = u v v u u^^v=u ox v-v ox u\boldsymbol{u} \wedge \boldsymbol{v}=\boldsymbol{u} \otimes \boldsymbol{v}-\boldsymbol{v} \otimes \boldsymbol{u}uv=uvvu can contribute:
(2) δ A α + 1 2 R α β γ δ A β ( u v ) γ δ = 0 (2) δ A α + 1 2 R α β γ δ A β ( u v ) γ δ = 0 {:(2)deltaA^(alpha)+(1)/(2)R^(alpha)_(beta gamma delta)A^(beta)(u^^v)^(gamma delta)=0:}\begin{equation*} \delta A^{\alpha}+\frac{1}{2} R^{\alpha}{ }_{\beta \gamma \delta} A^{\beta}(\boldsymbol{u} \wedge \boldsymbol{v})^{\gamma \delta}=0 \tag{2} \end{equation*}(2)δAα+12RαβγδAβ(uv)γδ=0
  1. This antisymmetric part is a "bivector." It is independent of the curve's shape; it depends only on (a) the plane the curve lies in, and (b) the area enclosed by the curve. [Although without metric "area" is meaningless, "relative areas at an event in a given plane" have just as much meaning as "relative lengths at an

Box 11.7 (continued)

event along a given direction." Two vectors at the same event lie on the same line if they are multiples of each other; their relative length in that case is their ratio. Similarly, two small closed curves at the same event lie in the same plane if their bivectors are multiples of each other (exercise 11.5); their relative area in that case is the ratio of their bivectors.]

B. General Case

Arbitrary but small closed curve.
  1. Break the curve down into a number of quadrilaterals, all lying in the same plane as the curve.
  2. Traverse each quadrilateral once in the same sense as the curve is to be traversed. Result: all interior edges get traversed twice in opposite directions (no net traversal); the outer edge (the
    curve itself) gets traversed once.
  3. Thus, δ A δ A delta A\delta \boldsymbol{A}δA due to traversing curve is the sum of the δ A δ A delta A\delta \boldsymbol{A}δA 's from traversal of each quadrilateral:
δ A α = 1 2 quadrilaterals R α β γ δ A β ( u v for given quadrilateral ) γ δ . δ A α = 1 2 quadrilaterals  R α β γ δ A β u v for given quadrilateral  γ δ . deltaA^(alpha)=-(1)/(2)sum_("quadrilaterals ")R^(alpha)_(beta gamma delta)A^(beta)(u^^v_("for given quadrilateral "))^(gamma delta).\delta A^{\alpha}=-\frac{1}{2} \sum_{\text {quadrilaterals }} R^{\alpha}{ }_{\beta \gamma \delta} A^{\beta}\left(\boldsymbol{u} \wedge \boldsymbol{v}_{\text {for given quadrilateral }}\right)^{\gamma \delta} .δAα=12quadrilaterals RαβγδAβ(uvfor given quadrilateral )γδ.
Define the bivector f f f\boldsymbol{f}f for the curve as the sum of the bivectors for its component quadrilaterals:
f quadrilaterals ( u v ) quadrilateral f quadrilaterals  ( u v ) quadrilateral  f-=sum_("quadrilaterals ")(u^^v)_("quadrilateral ")\boldsymbol{f} \equiv \sum_{\text {quadrilaterals }}(\mathbf{u} \wedge \boldsymbol{v})_{\text {quadrilateral }}fquadrilaterals (uv)quadrilateral 
(add "areas"; keep plane the same).
4. Then
δ A α + 1 2 R α β γ δ A β f γ δ δ A α + R α β | γ δ | A β f γ δ = 0 . δ A α + 1 2 R α β γ δ A β f γ δ δ A α + R α β | γ δ | A β f γ δ = 0 . deltaA^(alpha)+(1)/(2)R^(alpha)_(beta gamma delta)A^(beta)f^(gamma delta)-=deltaA^(alpha)+R^(alpha)_(beta|gamma delta|)A^(beta)f^(gamma delta)=0.\delta A^{\alpha}+\frac{1}{2} R^{\alpha}{ }_{\beta \gamma \delta} A^{\beta} f^{\gamma \delta} \equiv \delta A^{\alpha}+R^{\alpha}{ }_{\beta|\gamma \delta|} A^{\beta} f^{\gamma \delta}=0 .δAα+12RαβγδAβfγδδAα+Rαβ|γδ|Aβfγδ=0.

C. Warning

This is valid only for closed curves of small compass: δ A δ A delta A\delta \boldsymbol{A}δA doubles when the area doubles; but the error increases by a factor 2 3 / 2 [ δ A Δ a Δ b 2 3 / 2 [ δ A Δ a Δ b ∼2^(3//2)[delta A prop Delta a Delta b\sim 2^{3 / 2}[\delta \boldsymbol{A} \propto \Delta a \Delta b23/2[δAΔaΔb in calculation of §11.4; but error ( Δ a ) 2 Δ b ( Δ a ) 2 Δ b prop(Delta a)^(2)Delta b\propto(\Delta a)^{2} \Delta b(Δa)2Δb or Δ a ( Δ b ) 2 ] Δ a ( Δ b ) 2 {: Delta a(Delta b)^(2)]\left.\Delta a(\Delta b)^{2}\right]Δa(Δb)2].

§11.5. FLATNESS IS EQUIVALENT TO ZERO RIEMANN CURVATURE

To say that space or spacetime or any other manifold is flat is to say that there exists a coordinate system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)} in which all geodesics appear straight:
(11.16) x α ( λ ) = a α + b α λ . (11.16) x α ( λ ) = a α + b α λ . {:(11.16)x^(alpha)(lambda)=a^(alpha)+b^(alpha)lambda.:}\begin{equation*} x^{\alpha}(\lambda)=a^{\alpha}+b^{\alpha} \lambda . \tag{11.16} \end{equation*}(11.16)xα(λ)=aα+bαλ.
(Example: Lorentz spacetime of special relativity, where test bodies move on such straight lines.) They can appear so if and only if the connection coefficients in the geodesic equation
(11.17) d 2 x β d λ 2 + Γ β μ ν d x μ d λ d x ν d λ = 0 , (11.17) d 2 x β d λ 2 + Γ β μ ν d x μ d λ d x ν d λ = 0 , {:(11.17)(d^(2)x^(beta))/(dlambda^(2))+Gamma^(beta)_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)=0",":}\begin{equation*} \frac{d^{2} x^{\beta}}{d \lambda^{2}}+\Gamma^{\beta}{ }_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}=0, \tag{11.17} \end{equation*}(11.17)d2xβdλ2+Γβμνdxμdλdxνdλ=0,
expressed in the same coordinate system, all vanish:
(11.18) Γ β μ ν = 0 . (11.18) Γ β μ ν = 0 . {:(11.18)Gamma^(beta)_(mu nu)=0.:}\begin{equation*} \Gamma^{\beta}{ }_{\mu \nu}=0 . \tag{11.18} \end{equation*}(11.18)Γβμν=0.
From the vanishing of these connection coefficients, it follows immediately (equation 11.12) that all the components of the curvature tensor are zero:
(11.19) R β γ μ ν = 0 . (11.19) R β γ μ ν = 0 . {:(11.19)R^(beta)_(gamma mu nu)=0.:}\begin{equation*} R^{\beta}{ }_{\gamma \mu \nu}=0 . \tag{11.19} \end{equation*}(11.19)Rβγμν=0.
[Geometric restatement of ( 11.16 ) ( 11.18 ) ( 11.16 ) ( 11.18 ) (11.16)longrightarrow(11.18)longrightarrow(11.16) \longrightarrow(11.18) \longrightarrow(11.16)(11.18) (11.19): For all geodesics to be straight in a given coordinate system means that initially parallel geodesics preserve their separation; the geodesic deviation is zero; and therefore the curvature vanishes.]
Is the converse true? Does zero Riemann curvature imply the existence of a coordinate system in which all geodesics appear straight? Yes, as one sees by the following construction.
Transport a vector parallel to itself from P 0 P 0 P_(0)\mathscr{P}_{0}P0 to Q Q Q\mathscr{Q}Q, and then back from 2 2 2\mathscr{2}2 to P 0 P 0 P_(0)\mathscr{P}_{0}P0 along a slightly different route. It returns to its starting point with no alteration in magnitude or direction, because Riemann everywhere vanishes. Therefore parallel transport of a base vector e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ from P 0 P 0 P_(0)\mathscr{P}_{0}P0 to Q Q Q\mathscr{Q}Q yields at Q Q Q\mathscr{Q}Q a base vector e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ that is independent, both in magnitude and in direction, of the route of transportation (for routes obtainable one from the other by any continuous sequence of deformations). As for Q Q Q\mathcal{Q}Q, so for all points of the manifold; and as for the one base vector e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ, so for a complete set of base vectors ( μ = 0 , 1 , 2 , 3 μ = 0 , 1 , 2 , 3 mu=0,1,2,3\mu=0,1,2,3μ=0,1,2,3 ): Parallel transport of a basis { e α ( P 0 ) } e α P 0 {e_(alpha)(P_(0))}\left\{\boldsymbol{e}_{\alpha}\left(\mathscr{P}_{0}\right)\right\}{eα(P0)} yields everywhere a field of frames ("frame field"), each base vector of which suffers zero change (relative to the frame field) on parallel transport from any point to any nearby point: thus,
(11.20) e μ = 0 (11.20) e μ = 0 {:(11.20)grade_(mu)=0:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{e}_{\mu}=0 \tag{11.20} \end{equation*}(11.20)eμ=0
or
(11.21) ν e μ ( e γ e μ ) = 0 (11.21) ν e μ e γ e μ = 0 {:(11.21)grad_(nu)e_(mu)(-=grad_(e_(gamma))e_(mu))=0:}\begin{equation*} \boldsymbol{\nabla}_{\nu} \boldsymbol{e}_{\mu}\left(\equiv \boldsymbol{\nabla}_{\boldsymbol{e}_{\boldsymbol{\gamma}}} \boldsymbol{e}_{\mu}\right)=0 \tag{11.21} \end{equation*}(11.21)νeμ(eγeμ)=0
With the vanishing of these individual derivatives, there also vanishes the commutator of any two basis-vector fields:
(11.22) [ e μ , e ν ] = μ e ν ν e μ = 0 0 = 0 (11.22) e μ , e ν = μ e ν ν e μ = 0 0 = 0 {:(11.22)[e_(mu),e_(nu)]=grad_(mu)e_(nu)-grad_(nu)e_(mu)=0-0=0:}\begin{equation*} \left[\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right]=\boldsymbol{\nabla}_{\mu} \boldsymbol{e}_{\nu}-\boldsymbol{\nabla}_{\nu} \boldsymbol{e}_{\mu}=0-0=0 \tag{11.22} \end{equation*}(11.22)[eμ,eν]=μeννeμ=00=0
Flatness of a manifold defined
Flatness implies
Riemann = 0 = 0 =0=0=0
Proof that Riemann = 0 = 0 =0=0=0 implies flatness
The gap in the quadrilateral of Figure 11.2 (there read " e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ " for " u u u\boldsymbol{u}u," " e ν e ν e_(nu)\boldsymbol{e}_{\nu}eν " for " v v v\boldsymbol{v}v ") closes up completely. Thereupon one can introduce coordinates x μ x μ x^(mu)x^{\mu}xμ, each of which increases with a motion in the direction of the corresponding vector field; and with appropriate scaling of these coordinates, one can write
(11.23) e μ = x μ (11.23) e μ = x μ {:(11.23)e_(mu)=(del)/(delx^(mu)):}\begin{equation*} \boldsymbol{e}_{\mu}=\frac{\partial}{\partial x^{\mu}} \tag{11.23} \end{equation*}(11.23)eμ=xμ
(see exercise 9.9). With this coordinate basis in hand, one can employ the formula
(11.24) α e β = e μ Γ μ β α (11.24) α e β = e μ Γ μ β α {:(11.24)grad_(alpha)e_(beta)=e_(mu)Gamma^(mu)_(beta alpha):}\begin{equation*} \boldsymbol{\nabla}_{\alpha} \boldsymbol{e}_{\beta}=\boldsymbol{e}_{\mu} \Gamma^{\mu}{ }_{\beta \alpha} \tag{11.24} \end{equation*}(11.24)αeβ=eμΓμβα
to calculate the connection coefficients. From the vanishing of the quantities on the left, one concludes that all the connection coefficients on the right ("bending of geodesics") must be zero; so spacetime is indeed flat.
Summary: Spacetime is flat-i.e., there exist "flat coordinates" in which Γ μ α β = 0 Γ μ α β = 0 Gamma^(mu)_(alpha beta)=0\Gamma^{\mu}{ }_{\alpha \beta}=0Γμαβ=0 everywhere and geodesics are straight lines, x α ( λ ) = a α + b α λ x α ( λ ) = a α + b α λ x^(alpha)(lambda)=a^(alpha)+b^(alpha)lambdax^{\alpha}(\lambda)=a^{\alpha}+b^{\alpha} \lambdaxα(λ)=aα+bαλ-if and only if Riemann = 0 = 0 =0=0=0.
Note: In the spacetime of Einstein, which has a metric, one can choose { e μ ( P 0 ) } e μ P 0 {e_(mu)(P_(0))}\left\{\boldsymbol{e}_{\mu}\left(\mathscr{P}_{0}\right)\right\}{eμ(P0)} in the above argument to be orthonormal, e μ e ν = η μ ν e μ e ν = η μ ν e_(mu)*e_(nu)=eta_(mu nu)\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}=\eta_{\mu \nu}eμeν=ημν at P 0 P 0 P_(0)\mathscr{P}_{0}P0. The resulting field of frames will then be orthonormal everywhere, and the resulting coordinate system will be Lorentz. Thus, in Einsteinian gravity the above summary can be rewritten: spacetime is flat (there exists a Lorentz coordinate system) if and only if Riemann = 0 = 0 =0=0=0.
Warning: Flatness does not necessarily imply Euclidean topology. Take a sheet of paper. It is flat. Roll it up into a cylinder. It is still flat, intrinsically. The tracks of geodesics over it have not changed. Distances between neighboring points have not changed. Only the topology has changed, so far as an observer confined forever to the sheet is concerned. (The "extrinsic geometry"-the way the sheet is embedded in the surrounding three-dimensional space-has also changed; but an observer on the sheet knows nothing of this, and it is not the subject of the present chapter. See, instead §21.5.)
Take this cylinder. Bend it around and glue its two ends together, without changing its flat intrinsic geometry. Doing so is impossible if the cylinder remains embedded in flat, three-dimensional Euclidean space; perfectly possible if it is embedded in a Euclidean space of 4 dimensions. However, embedding is unimportant to observers confined to the cylinder, since all they ever measure is intrinsic geometry; so all that matters to them is the topological identification of the two ends of the cylinder with each other. The result is topologically a torus; but the tracks of geodesics are still unchanged; the intrinsic geometry is flat; Riemann vanishes.
By analogy, take flat Minkowskii spacetime. Pick some Lorentz frame, and in it pick a cube 10 10 10 10 10^(10)10^{10}1010 light years on each side ( 0 < x < 10 10 0 < x < 10 10 (0 < x < 10^(10):}\left(0<x<10^{10}\right.(0<x<1010 light years; similarly for y y yyy and z z zzz ). Identify opposite faces of the cube so that a geodesic exiting across one face enters across the other. The result is topologically a three-torus: a "closed universe" with finite volume, with flat, Minkowskii geometry, and with a form that changes not at all as Lorentz time t t ttt passes (no expansion, no contraction).

§11.6. RIEMANN NORMAL COORDINATES

In curved spacetime one can never find a coordinate system with Γ α β γ = 0 Γ α β γ = 0 Gamma^(alpha)_(beta gamma)=0\Gamma^{\alpha}{ }_{\beta \gamma}=0Γαβγ=0 everywhere. But one can always construct local inertial frames at a given event P 0 P 0 P_(0)\mathscr{P}_{0}P0; and as viewed in such frames, free particles must move along straight lines, at least locally-which means Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ must vanish, at least locally.
A very special and useful realization of such a local inertial frame is a Riemannnormal coordinate system. Pick an event P 0 P 0 P_(0)\mathscr{P}_{0}P0 and a set of basis vectors { e α ( P 0 ) } e α P 0 {e_(alpha)(P_(0))}\left\{\boldsymbol{e}_{\alpha}\left(\mathscr{P}_{0}\right)\right\}{eα(P0)} to be used there by an inertial observer. Fill spacetime, near P 0 P 0 P_(0)\mathscr{P}_{0}P0, with geodesics radiating out from P 0 P 0 P_(0)\mathscr{P}_{0}P0 like the quills of a hedgehog or porcupine. Each geodesic is determined by its tangent vector v v v\boldsymbol{v}v at P 0 P 0 P_(0)\mathscr{P}_{0}P0; and the general point on it can be denoted
(11.25) P = G ( λ ; V ) . [ affine parameter; tells "where" on geodesic ] tangent vector at P 0 ; tells "which geodesic" } (11.25) P = G ( λ ; V ) .  affine parameter;   tells "where" on geodesic   tangent vector at  P 0 ;  tells "which geodesic"  {:(11.25){:[P=G(lambda;V).],[[[" affine parameter; "],[" tells "where" on geodesic "]]][" tangent vector at "P_(0);],[" tells "which geodesic" "]}:}\left.\begin{array}{l} \mathscr{P}=\mathscr{G}(\lambda ; \boldsymbol{V}) . \tag{11.25}\\ {\left[\begin{array}{l} \text { affine parameter; } \\ \text { tells "where" on geodesic } \end{array}\right]} \end{array} \begin{array}{l} \text { tangent vector at } \mathscr{P}_{0} ; \\ \text { tells "which geodesic" } \end{array}\right\}(11.25)P=G(λ;V).[ affine parameter;  tells "where" on geodesic ] tangent vector at P0; tells "which geodesic" }
Actually, this gives more geodesics than are needed. One reaches the same point after parameter length 1 2 λ 1 2 λ (1)/(2)lambda\frac{1}{2} \lambda12λ if the initial tangent vector is 2 v 2 v 2v2 \boldsymbol{v}2v, as one reaches after λ λ lambda\lambdaλ if the tangent vector is v v v\boldsymbol{v}v :
G ( λ ; v ) = G ( 1 2 λ ; 2 v ) = G ( 1 ; λ v ) G ( λ ; v ) = G 1 2 λ ; 2 v = G ( 1 ; λ v ) G(lambda;v)=G((1)/(2)lambda;2v)=G(1;lambda v)\mathscr{G}(\lambda ; \boldsymbol{v})=\mathscr{G}\left(\frac{1}{2} \lambda ; 2 \boldsymbol{v}\right)=\mathscr{G}(1 ; \lambda \boldsymbol{v})G(λ;v)=G(12λ;2v)=G(1;λv)
Thus, by fixing λ = 1 λ = 1 lambda=1\lambda=1λ=1 and varying v v v\boldsymbol{v}v in all possible ways, one can reach every point in some neighborhood of P 0 P 0 P_(0)\mathscr{P}_{0}P0. This is the foundation for constructing Riemann normal coordinates. Choose an event P P P\mathscr{P}P. Find that tangent vector v v v\boldsymbol{v}v at P 0 P 0 P_(0)\mathscr{P}_{0}P0 for which P = P = P=\mathscr{P}=P= G ( 1 ; v ) G ( 1 ; v ) G(1;v)\mathscr{G}(1 ; \boldsymbol{v})G(1;v). Expand that v v v\boldsymbol{v}v in terms of the chosen basis and give its components the names x α x α x^(alpha)x^{\alpha}xα :
(11.26) P = G ( 1 ; x α e α ) . (11.26) P = G 1 ; x α e α . {:(11.26)P=G(1;x^(alpha)e_(alpha)).:}\begin{equation*} \mathscr{P}=\mathscr{G}\left(1 ; x^{\alpha} \boldsymbol{e}_{\alpha}\right) . \tag{11.26} \end{equation*}(11.26)P=G(1;xαeα).
The point P P P\mathscr{P}P determines x α x α x^(alpha)x^{\alpha}xα uniquely (if P P P\mathscr{P}P is near enough to P 0 P 0 P_(0)\mathscr{P}_{0}P0 that spacetime curvature has not caused geodesics to cross each other). Similarly, x α x α x^(alpha)x^{\alpha}xα determines P P P\mathscr{P}P uniquely. Hence, x α x α x^(alpha)x^{\alpha}xα can be chosen as the coordinates of P P P\mathscr{P}P-its "Riemann-normal coordinates, based on the event P 0 P 0 P_(0)\mathscr{P}_{0}P0 and basis { e α ( P 0 ) } . " e α P 0 . " {e_(alpha)(P_(0))}."\left\{\boldsymbol{e}_{\alpha}\left(\mathscr{P}_{0}\right)\right\} . "{eα(P0)}."
Equation (11.26) summarizes Riemann-normal coordinates concisely. Other equations, derived in exercise 11.9, summarize their powerful properties:
(11.27) e α ( P 0 ) = ( / x α ) P 0 ; (11.28) Γ α β γ ( P 0 ) = 0 ; (11.29) Γ α β γ , μ ( P 0 ) = 1 3 ( R α β γ μ + R α γ β μ ) . (11.27) e α P 0 = / x α P 0 ; (11.28) Γ α β γ P 0 = 0 ; (11.29) Γ α β γ , μ P 0 = 1 3 R α β γ μ + R α γ β μ . {:[(11.27)e_(alpha)(P_(0))=(del//delx^(alpha))_(P_(0));],[(11.28)Gamma^(alpha)_(beta gamma)(P_(0))=0;],[(11.29)Gamma^(alpha)_(beta gamma,mu)(P_(0))=-(1)/(3)(R^(alpha)_(beta gamma mu)+R^(alpha)_(gamma beta mu)).]:}\begin{align*} \boldsymbol{e}_{\alpha}\left(\mathscr{P}_{0}\right) & =\left(\partial / \partial x^{\alpha}\right)_{\mathscr{P}_{0}} ; \tag{11.27}\\ \Gamma^{\alpha}{ }_{\beta \gamma}\left(\mathscr{P}_{0}\right) & =0 ; \tag{11.28}\\ \Gamma^{\alpha}{ }_{\beta \gamma, \mu}\left(\mathscr{P}_{0}\right) & =-\frac{1}{3}\left(R^{\alpha}{ }_{\beta \gamma \mu}+R^{\alpha}{ }_{\gamma \beta \mu}\right) . \tag{11.29} \end{align*}(11.27)eα(P0)=(/xα)P0;(11.28)Γαβγ(P0)=0;(11.29)Γαβγ,μ(P0)=13(Rαβγμ+Rαγβμ).
If spacetime has a metric (as it does in actuality), and if the observer's frame at P 0 P 0 P_(0)\mathscr{P}_{0}P0 has been chosen orthonormal ( e α e β = η α β ) e α e β = η α β (e_(alpha)*e_(beta)=eta_(alpha beta))\left(\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}=\eta_{\alpha \beta}\right)(eαeβ=ηαβ), then
Riemann normal coordinates: a realization of local inertial frames
Geometric construction of Riemann normal coordinates
(11.30) g α β ( P 0 ) = η α β , (11.31) g α β , μ ( P 0 ) = 0 , (11.32) g α β , μ v ( P 0 ) = 1 3 ( R α μ β v + R α ν β μ ) = 2 3 J α β μ ν , (11.32') R α β γ δ ( P 0 ) = g α δ , β γ ( P 0 ) g α γ , β δ ( P 0 ) . (11.30) g α β P 0 = η α β , (11.31) g α β , μ P 0 = 0 , (11.32) g α β , μ v P 0 = 1 3 R α μ β v + R α ν β μ = 2 3 J α β μ ν , (11.32') R α β γ δ P 0 = g α δ , β γ P 0 g α γ , β δ P 0 . {:[(11.30)g_(alpha beta)(P_(0))=eta_(alpha beta)","],[(11.31)g_(alpha beta,mu)(P_(0))=0","],[(11.32)g_(alpha beta,mu v)(P_(0))=-(1)/(3)(R_(alpha mu beta v)+R_(alpha nu beta mu))],[=-(2)/(3)J_(alpha beta mu nu)","],[(11.32')R_(alpha beta gamma delta)(P_(0))=g_(alpha delta,beta gamma)(P_(0))-g_(alpha gamma,beta delta)(P_(0)).]:}\begin{align*} g_{\alpha \beta}\left(\mathscr{P}_{0}\right) & =\eta_{\alpha \beta}, \tag{11.30}\\ g_{\alpha \beta, \mu}\left(\mathscr{P}_{0}\right) & =0, \tag{11.31}\\ g_{\alpha \beta, \mu v}\left(\mathscr{P}_{0}\right) & =-\frac{1}{3}\left(R_{\alpha \mu \beta v}+R_{\alpha \nu \beta \mu}\right) \tag{11.32}\\ & =-\frac{2}{3} J_{\alpha \beta \mu \nu}, \\ R_{\alpha \beta \gamma \delta}\left(\mathscr{P}_{0}\right) & =g_{\alpha \delta, \beta \gamma}\left(\mathscr{P}_{0}\right)-g_{\alpha \gamma, \beta \delta}\left(\mathscr{P}_{0}\right) . \tag{11.32'} \end{align*}(11.30)gαβ(P0)=ηαβ,(11.31)gαβ,μ(P0)=0,(11.32)gαβ,μv(P0)=13(Rαμβv+Rανβμ)=23Jαβμν,(11.32')Rαβγδ(P0)=gαδ,βγ(P0)gαγ,βδ(P0).
Here J α β μ ν J α β μ ν J_(alpha beta mu nu)J_{\alpha \beta \mu \nu}Jαβμν are components of the Jacobi curvature tensor (see exercise 11.7).
Is this the only coordinate system that is locally inertial at P 0 P 0 P_(0)\mathscr{P}_{0}P0 (i.e., has Γ α β γ = 0 Γ α β γ = 0 Gamma^(alpha)_(beta gamma)=0\Gamma^{\alpha}{ }_{\beta \gamma}=0Γαβγ=0
Other mathematical realizations of a local inertial frame
there) and is tied to the basis vectors e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα there (i.e., has / x α = e α / x α = e α del//delx^(alpha)=e_(alpha)\partial / \partial x^{\alpha}=\boldsymbol{e}_{\alpha}/xα=eα there)? No. But all such coordinate systems (called "normal coordinates") will be the same to second order:
x NEW α ( P ) = x OLD α ( P ) + corrections of order ( x OLD α ) 3 . x NEW α ( P ) = x OLD α ( P ) +  corrections of order  x OLD α 3 . x_(NEW)^(alpha)(P)=x_(OLD)^(alpha)(P)+" corrections of order "(x_(OLD)^(alpha))^(3).x_{\mathrm{NEW}}^{\alpha}(\mathscr{P})=x_{\mathrm{OLD}}^{\alpha}(\mathscr{P})+\text { corrections of order }\left(x_{\mathrm{OLD}}^{\alpha}\right)^{3} .xNEWα(P)=xOLDα(P)+ corrections of order (xOLDα)3.
Moreover, only those the same to third order,
x NEW α ( P ) = x OLD α ( P ) + corrections of order ( x OLD α ) 4 , x NEW α ( P ) = x OLD α ( P ) +  corrections of order  x OLD α 4 , x_(NEW)^(alpha)(P)=x_(OLD)^(alpha)(P)+" corrections of order "(x_(OLD)^(alpha))^(4),x_{\mathrm{NEW}}^{\alpha}(\mathscr{P})=x_{\mathrm{OLD}}^{\alpha}(\mathscr{P})+\text { corrections of order }\left(x_{\mathrm{OLD}}^{\alpha}\right)^{4},xNEWα(P)=xOLDα(P)+ corrections of order (xOLDα)4,
will preserve the beautiful ties (11.29) and (11.32) to the Riemann curvature tensor.

EXERCISES

Exercise 11.6. SYMMETRIES OF Riemann

(To be discussed in Chapter 13). Show that Riemann has the following symmetries:
(11.33a) R α β γ δ = R α β [ γ δ ] (antisymmetric on last 2 indices) (11.33b) R α [ β γ δ ] = 0 (vanishing of completely antisymmetric part) (11.33a) R α β γ δ = R α β [ γ δ ]  (antisymmetric on last 2 indices)  (11.33b) R α [ β γ δ ] = 0  (vanishing of completely antisymmetric part)  {:[(11.33a)R^(alpha)_(beta gamma delta)=R^(alpha)_(beta[gamma delta])" (antisymmetric on last 2 indices) "],[(11.33b)R^(alpha)_([beta gamma delta])=0" (vanishing of completely antisymmetric part) "]:}\begin{align*} R^{\alpha}{ }_{\beta \gamma \delta} & =R^{\alpha}{ }_{\beta[\gamma \delta]} & & \text { (antisymmetric on last 2 indices) } \tag{11.33a}\\ R^{\alpha}{ }_{[\beta \gamma \delta]} & =0 & & \text { (vanishing of completely antisymmetric part) } \tag{11.33b} \end{align*}(11.33a)Rαβγδ=Rαβ[γδ] (antisymmetric on last 2 indices) (11.33b)Rα[βγδ]=0 (vanishing of completely antisymmetric part) 

Exercise 11.7. GEODESIC DEVIATION MEASURES ALL CURVATURE COMPONENTS

The equation of geodesic deviation, written up to now in the form
or
u u n + Riemann ( , u , n , u ) = 0 u u n + ( n , u ) u = 0 u u n + Riemann ( , u , n , u ) = 0 u u n + ( n , u ) u = 0 {:[grad_(u)grad_(u)n+Riemann(dots","u","n","u)=0],[grad_(u)grad_(u)n+ℜ(n","u)u=0]:}\begin{gathered} \boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} n+\operatorname{Riemann}(\ldots, u, n, u)=0 \\ \boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} n+\Re(n, u) u=0 \end{gathered}uun+Riemann(,u,n,u)=0uun+(n,u)u=0
also lets itself be written in the Jacobi form u u n + f ( u , u ) n = 0 u u n + f ( u , u ) n = 0 grad_(u)grad_(u)n+f(u,u)n=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}+\mathscr{f}(\boldsymbol{u}, \boldsymbol{u}) \boldsymbol{n}=0uun+f(u,u)n=0. Here G ( u , v ) G ( u , v ) G(u,v)\mathscr{G}(\boldsymbol{u}, \boldsymbol{v})G(u,v), the "Jacobi curvature operator," is defined by
(11.34) f ( u , v ) n 1 2 [ R ( n , u ) v + R ( n , v ) u ] , (11.34) f ( u , v ) n 1 2 [ R ( n , u ) v + R ( n , v ) u ] , {:(11.34)f(u","v)n-=(1)/(2)[R(n","u)v+R(n","v)u]",":}\begin{equation*} \mathscr{f}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{n} \equiv \frac{1}{2}[\mathscr{R}(\boldsymbol{n}, \boldsymbol{u}) \boldsymbol{v}+\mathscr{R}(\boldsymbol{n}, \boldsymbol{v}) \boldsymbol{u}], \tag{11.34} \end{equation*}(11.34)f(u,v)n12[R(n,u)v+R(n,v)u],
and is related to the "Jacobi curvature tensor" by
(11.35) Jacobi ( , n , u , v ) f ( u , v ) n (11.35) Jacobi ( , n , u , v ) f ( u , v ) n {:(11.35)Jacobi(dots","n","u","v)-=f(u","v)n:}\begin{equation*} \operatorname{Jacobi}(\ldots, n, u, v) \equiv \mathscr{f}(u, v) n \tag{11.35} \end{equation*}(11.35)Jacobi(,n,u,v)f(u,v)n
which implies
(11.36) J μ ν α β = J μ β α = 1 2 ( R α ν β μ + R β ν α μ ) . (11.36) J μ ν α β = J μ β α = 1 2 R α ν β μ + R β ν α μ . {:(11.36)J^(mu)^(nu alpha beta)^(')=J^(mu beta alpha)=(1)/(2)(R_(alpha nu beta)^(mu)+R_(beta nu alpha)^(mu)).:}\begin{equation*} {J^{\mu}}^{\nu \alpha \beta}{ }^{\prime}={J^{\mu \beta \alpha}}=\frac{1}{2}\left(R_{\alpha \nu \beta}^{\mu}+R_{\beta \nu \alpha}^{\mu}\right) . \tag{11.36} \end{equation*}(11.36)Jμναβ=Jμβα=12(Rανβμ+Rβναμ).
(a) Show that J μ ( α β γ ) = 0 J μ ( α β γ ) = 0 J^(mu)_((alpha beta gamma))=0J^{\mu}{ }_{(\alpha \beta \gamma)}=0Jμ(αβγ)=0 follows from R μ α β γ = R μ α [ β γ ] R μ α β γ = R μ α [ β γ ] R^(mu)_(alpha beta gamma)=R^(mu)_(alpha[beta gamma])R^{\mu}{ }_{\alpha \beta \gamma}=R^{\mu}{ }_{\alpha[\beta \gamma]}Rμαβγ=Rμα[βγ].
(b) Show that by studying geodesic deviation (allowing arbitrary u u u\boldsymbol{u}u and n n n\boldsymbol{n}n in u u n + u u n + grad_(u)grad_(u)n+\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}+uun+ f ( u , u ) n = 0 f ( u , u ) n = 0 f(u,u)n=0\mathscr{f}(\boldsymbol{u}, \boldsymbol{u}) \boldsymbol{n}=0f(u,u)n=0 ) one can measure all components of Jacobi.
(c) Show that Jacobi contains precisely the same information as Riemann. [Hint: show that
(11.37) R α ν β μ = 2 3 ( J μ ν α β J μ β α ν ) ; (11.37) R α ν β μ = 2 3 J μ ν α β J μ β α ν ; {:(11.37)R_(alpha nu beta)^(mu)=(2)/(3)(J^(mu)_(nu alpha beta)-J^(mu)_(beta alpha nu));:}\begin{equation*} R_{\alpha \nu \beta}^{\mu}=\frac{2}{3}\left(J^{\mu}{ }_{\nu \alpha \beta}-J^{\mu}{ }_{\beta \alpha \nu}\right) ; \tag{11.37} \end{equation*}(11.37)Rανβμ=23(JμναβJμβαν);
this plus equation (11.36) for J μ v α β J μ v α β J^(mu)_(v alpha beta)J^{\mu}{ }_{v \alpha \beta}Jμvαβ proves "same information content".] Hence, by studying geodesic deviation one can also measure all the components of Riemann.
(d) Show that the symmetry of R [ ν α β ] μ = 0 R [ ν α β ] μ = 0 R_([nu alpha beta])^(mu)=0R_{[\nu \alpha \beta]}^{\mu}=0R[ναβ]μ=0 is essential in the equivalence between Jacobi and Riemann by exhibiting proposed values for R μ ν α β = R μ ν β α R μ ν α β = R μ ν β α R^(mu)_(nu alpha beta)=-R^(mu)_(nu beta alpha)R^{\mu}{ }_{\nu \alpha \beta}=-R^{\mu}{ }_{\nu \beta \alpha}Rμναβ=Rμνβα for which R μ [ ν α β ] 0 R μ [ ν α β ] 0 R^(mu)_([nu alpha beta])!=0R^{\mu}{ }_{[\nu \alpha \beta]} \neq 0Rμ[ναβ]0, and from which one would find J μ ν α β = 0 J μ ν α β = 0 J^(mu)^(nu alpha beta)=0{J^{\mu}}^{\nu \alpha \beta}=0Jμναβ=0.

Exercise 11.8. GEODESIC DEVIATION IN GORY DETAIL

Write out the equation of geodesic deviation in component form in a coordinate system. Expand all covariant derivatives (semicolon notation) in terms of ordinary (comma) derivatives and in terms of Γ Γ Gamma\GammaΓ 's to show all Γ Γ Gamma\GammaΓ and del\partial terms explicitly.

Exercise 11.9. RIEMANN NORMAL COORDINATES IN GENERAL

Derive properties (11.27), (11.28), (11.29), (11.31), (11.32), and (11.32') of Riemann normal coordinates. Hint: Proceed as follows.
(a) From definition (11.26), derive ( P / x α ) F 0 = e α P / x α F 0 = e α (delP//delx^(alpha))_(F_(0))=e_(alpha)\left(\partial \mathscr{P} / \partial x^{\alpha}\right)_{\mathscr{F}_{0}}=\boldsymbol{e}_{\alpha}(P/xα)F0=eα.
(b) Similarly, from definition (11.26), show that each of the curves x α = v α λ x α = v α λ x^(alpha)=v^(alpha)lambdax^{\alpha}=v^{\alpha} \lambdaxα=vαλ (where the v α v α v^(alpha)v^{\alpha}vα are constants) is a geodesic through P 0 P 0 P_(0)\mathscr{P}_{0}P0, with affine parameter λ λ lambda\lambdaλ.
(c) Show that Γ α β γ ( P 0 ) = 0 Γ α β γ P 0 = 0 Gamma^(alpha)_(beta gamma)(P_(0))=0\Gamma^{\alpha}{ }_{\beta \gamma}\left(\mathscr{P}_{0}\right)=0Γαβγ(P0)=0 by substituting x α = v α λ x α = v α λ x^(alpha)=v^(alpha)lambdax^{\alpha}=v^{\alpha} \lambdaxα=vαλ into the geodesic equation.
(d) Since the curves x α = v α λ x α = v α λ x^(alpha)=v^(alpha)lambdax^{\alpha}=v^{\alpha} \lambdaxα=vαλ are geodesics for every choice of the parameters v α v α v^(alpha)v^{\alpha}vα, they provide not only a geodesic tangent u ( / λ ) v u ( / λ ) v u-=(del//del lambda)_(v^(**))\boldsymbol{u} \equiv(\partial / \partial \lambda)_{v^{*}}u(/λ)v, but also several deviation vectors N ( α ) ( / v α ) λ N ( α ) / v α λ N_((alpha))-=(del//delv^(alpha))_(lambda)\boldsymbol{N}_{(\alpha)} \equiv\left(\partial / \partial v^{\alpha}\right)_{\lambda}N(α)(/vα)λ. Compute the components of these vectors in the Riemann normal coordinate system, and substitute into the geodesic deviation equation as written in exercise 11.8 .
(e) Equate to zero the coefficients of the zeroth and first powers of λ λ lambda\lambdaλ in the geodesic deviation equation of part (d), using
Γ β γ α | z μ = v n λ = λ v μ Γ β γ , μ α ( P 0 ) + O ( λ 2 ) , Γ β γ α z μ = v n λ = λ v μ Γ β γ , μ α P 0 + O λ 2 , Gamma_(beta gamma)^(alpha)|_(z^(mu)=v^(n lambda))=lambdav^(mu)Gamma_(beta gamma,mu)^(alpha)(P_(0))+O(lambda^(2)),\left.\Gamma_{\beta \gamma}^{\alpha}\right|_{z^{\mu}=v^{n \lambda}}=\lambda v^{\mu} \Gamma_{\beta \gamma, \mu}^{\alpha}\left(\mathscr{P}_{0}\right)+O\left(\lambda^{2}\right),Γβγα|zμ=vnλ=λvμΓβγ,μα(P0)+O(λ2),
which is a Taylor series for Γ Γ Gamma\GammaΓ. In this way arrive at equation (11.29) for Γ α β γ , μ Γ α β γ , μ Gamma^(alpha)_(beta gamma,mu)\Gamma^{\alpha}{ }_{\beta \gamma, \mu}Γαβγ,μ in terms of the Riemann tensor.
(f) From equations (11.28), (11.29), and (8.24) for the connection coefficients in terms of the metric, derive equations (11.31), (11.32), and (11.32').

Exercise 11.10. BIANCHI IDENTITIES

Show that the Riemann curvature tensor satisfies the following "Bianchi identities"
(11.38) R β γ δ ; ϵ ] α = 0 (11.38) R β γ δ ; ϵ ] α = 0 {:(11.38)R_(beta|__ gamma delta;epsilon])^(alpha)=0:}\begin{equation*} R_{\beta\lfloor\gamma \delta ; \epsilon]}^{\alpha}=0 \tag{11.38} \end{equation*}(11.38)Rβγδ;ϵ]α=0
The geometric meaning of these identities will be discussed in Chapter 15. [Hint: Perform the calculation at the origin of a Riemann normal coordinate system.]

Exercise 11.11. CURVATURE OPERATOR ACTS ON 1-FORMS

Let R ( u , v ) R ( u , v ) R(u,v)\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})R(u,v) be the operator R ( u , v ) = [ u , v ] [ u , v ] R ( u , v ) = u , v [ u , v ] R(u,v)=[grad_(u),grad_(v)]-grad_([u,v])\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})=\left[\boldsymbol{\nabla}_{\boldsymbol{u}}, \boldsymbol{\nabla}_{\boldsymbol{v}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{u}, \boldsymbol{v}]}R(u,v)=[u,v][u,v] when acting on 1-forms σ σ sigma\boldsymbol{\sigma}σ (or other tensors) as well as on tangent vectors. Show that
R ( u , v ) σ , w = σ , R ( u , v ) w . R ( u , v ) σ , w = σ , R ( u , v ) w . (:R(u,v)sigma,w:)=-(:sigma,R(u,v)w:).\langle\mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{\sigma}, \boldsymbol{w}\rangle=-\langle\boldsymbol{\sigma}, \mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{w}\rangle .R(u,v)σ,w=σ,R(u,v)w.

Exercise 11.12. ROTATION GROUP: RIEMANN CURVATURE

[Continuation of exercises 9.13, 9.14, and 10.17.] Calculate the components of the Riemann curvature tensor for the rotation group's manifold S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3); use the basis of generators { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα}. [Answer:
(11.39) R α β γ δ = 1 2 δ γ δ α β , (11.39) R α β γ δ = 1 2 δ γ δ α β , {:(11.39)R^(alpha)_(beta gamma delta)=(1)/(2)delta_(gamma delta)^(alpha beta)",":}\begin{equation*} R^{\alpha}{ }_{\beta \gamma \delta}=\frac{1}{2} \delta_{\gamma \delta}^{\alpha \beta}, \tag{11.39} \end{equation*}(11.39)Rαβγδ=12δγδαβ,
where δ γ δ α β δ γ δ α β delta_(gamma delta)^(alpha beta)\delta_{\gamma \delta}^{\alpha \beta}δγδαβ is the permutation symbol defined in equation (3.501):
δ γ δ α β ( δ α γ δ β δ δ α δ δ β γ ) . δ γ δ α β δ α γ δ β δ δ α δ δ β γ . delta_(gamma delta)^(alpha beta)-=(delta^(alpha)_(gamma)delta^(beta)_(delta)-delta^(alpha)_(delta)delta^(beta)_(gamma)).\delta_{\gamma \delta}^{\alpha \beta} \equiv\left(\delta^{\alpha}{ }_{\gamma} \delta^{\beta}{ }_{\delta}-\delta^{\alpha}{ }_{\delta} \delta^{\beta}{ }_{\gamma}\right) .δγδαβ(δαγδβδδαδδβγ).
Note that this answer is independent of location P P P\mathscr{P}P in the group manifold.]

  1. *If metric is absent, these starred formulas cannot be formulated. All other formulas are valid in absence of metric.